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Indices of pentagonal numbers > 0 which are not the difference of two larger pentagonal numbers.
+10
12
1, 2, 3, 5, 6, 8, 9, 11, 15, 18, 23, 24, 27, 51, 54, 71, 72, 81, 96, 99, 123, 128, 135, 162, 216, 239, 243, 263, 288, 303, 311, 359, 384, 423, 459, 479, 486, 519, 591, 599, 639, 648, 683, 699, 729, 743, 783, 863, 864, 879, 891, 911, 1031, 1103, 1151, 1215, 1431
EXAMPLE
a(1..3)=1,2,3 since P(1),P(2),P(3) cannot be written as difference of 2 other pentagonal numbers.
P(4)=22=P(8)-P(7), therefore 4 is not in this sequence.
PROG
(PARI) P(n)=n*(3*n-1)>>1
isPent(t)=P(sqrtint((t<<1)\3)+1)==t
for( i=1, 999, for( j=i+1, (P(i)-1)\3, isPent(P(i)+P(j))&next(2)); print1(i", "))
EXTENSIONS
Definition corrected, thanks to a remark from R. J. Mathar, Feb 01 2008
Pentagonal numbers > 0 which are not the difference of two larger pentagonal numbers.
+10
6
1, 5, 12, 35, 51, 92, 117, 176, 330, 477, 782, 852, 1080, 3876, 4347, 7526, 7740, 9801, 13776, 14652, 22632, 24512, 27270, 39285, 69876, 85562, 88452, 103622, 124272, 137562, 144926, 193142, 220992, 268182, 315792, 343922, 354051, 403782, 523626
EXAMPLE
a(1..3)=P(1),P(2),P(3) since these cannot be written as difference of 2 other pentagonal numbers > 0.
P(4)=22=P(8)-P(7), therefore P(4) is not in this sequence.
PROG
(PARI) P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */ isPent(t)=P(sqrtint((t<<1)\3)+1)==t
for( i=1, 999, for( j=i+1, (P(i)-1)\3, isPent(P(i)+P(j))&next(2)); print1(P(i)", "))
Pentagonal numbers ( A000326) which are the sum of 2 other positive pentagonal numbers.
+10
6
70, 92, 852, 925, 1247, 1426, 1926, 2625, 3577, 5192, 6305, 6501, 7107, 7740, 7957, 8177, 8626, 9560, 10292, 12927, 13207, 14652, 15555, 16172, 18095, 20475, 20827, 21901, 22265, 22632, 23002, 23751, 24130, 28497, 29330, 31032, 33227, 33675
COMMENTS
It is conjectured that every integer and hence every pentagonal number, greater than 33066, hence greater than A000326(149) = 33227, can be represented as the sum of three pentagonal numbers. - Jonathan Vos Post, Dec 18 2007
EXAMPLE
a(1)=70=P(7) is the least pentagonal number which can be written as sum of two other pentagonal numbers, P(7)=P(5)+P(5).
PROG
(PARI) P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */ isPent(t)=P(sqrtint(t<<1\3)+1)==t
for(i=1, 299, for(j=1, (i+1)\sqrt(2), isPent(P(i)-P(j)) && print1(P(i)", ") || next(2)))
/* The following is much faster, at the cost of implementing sum2sqr(), cf. A133388*/ A136117next(i)=i=sqrtint(i\3*2)*6+5; until(0, for(j=2, #t=sum2sqr((i+=6)^2+1), t[j]%6==[5, 5] && break(2))); i^2\24
A136117vect(n, i)=vector(n, j, i=A136117next(i)) /* 2nd arg =0 by default but allows one to start elsewhere */
Least pentagonal number P(m) > P(n) such that P(m)+P(n) is again a pentagonal number, 0 if no such m exists.
+10
4
0, 0, 0, 70, 0, 0, 782, 0, 0, 3432, 0, 715, 10045, 3290, 0, 23375, 1001, 0, 715, 2035, 7526, 84847, 0, 0, 1001, 41251, 0, 4030, 64170, 32340, 20475, 14652, 11310, 490490, 7740, 6700, 5922, 190995, 4845, 18095, 259376, 3876, 1260875, 10045, 3290
EXAMPLE
a(1..3)=0 since P(1),P(2),P(3) cannot be written as difference of 2 other pentagonal numbers > 0.
a(4)=70=P(7) is the least pentagonal number > P(4)=22 such that their sum is again a pentagonal number, P(8).
PROG
(PARI) P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */ /* newline */ isPent(t)=P(sqrtint(t<<1\3)+1)==t /* newline */ for( i=1, 99, for( j=i+1, (P(i)-1)\3, isPent(P(i)+P(j))&print1(P(j)", ")|next(2)); print1(0", "))
Indices of pentagonal numbers ( A000326) which are the sum of 2 other positive pentagonal numbers.
+10
4
7, 8, 24, 25, 29, 31, 36, 42, 49, 59, 65, 66, 69, 72, 73, 74, 76, 80, 83, 93, 94, 99, 102, 104, 110, 117, 118, 121, 122, 123, 124, 126, 127, 138, 140, 144, 149, 150, 152, 161, 163, 168, 169, 174, 175, 178, 181, 185, 188, 190, 195, 199, 203, 209, 210, 212, 213
EXAMPLE
a(1)=7 since P(7)=70 is the least pentagonal number which can be written as sum of two other pentagonal numbers, P(7)=P(5)+P(5).
PROG
(PARI) P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */ isPent(t)=P(sqrtint(t<<1\3)+1)==t
for(i=1, 999, for(j=1, (i+1)\sqrt(2), isPent(P(i)-P(j))&print1(i", ") || next(2)))
/* The following are much faster, at the cost of implementing sum2sqr(), cf. A133388. */ A136116next(i)=i=6*i-1; until(0, for(j=2, #t=sum2sqr((i+=6)^2+1), t[j]%6==[5, 5] && break(2))); i\6+1
A136116vect(n, i=0)=vector(n, j, i=A136116next(i))
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