cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 11 results. Next

A135771 Terms in A136112 which are not in A135768.

Original entry on oeis.org

5, 23, 51, 71, 72, 99, 123, 239, 263, 311, 359, 479, 599, 699, 743, 863, 911, 1031, 1103, 1151, 1431, 1563, 1583, 1823, 1851, 1863, 2111, 2543, 2663, 3023, 3119, 3191, 3291, 3671, 3719, 3863, 4131, 4203, 4271, 4463, 4671, 4703, 5039, 5231, 5351, 5391, 5399
Offset: 1

Views

Author

R. J. Mathar and M. F. Hasler, Feb 07 2008

Keywords

Comments

Pentagonal-Indices of terms in A136113 which are not in A135769.
A135768 resp. A135769 are subsequences of A136112 resp. A136113; the present sequence gives the indices of the elements of the former which are not in the latter: A136113(A135771(k)), k=1,2,3,... are the pentagonal numbers P(m) which are not the difference of two pentagonal numbers P(n)-P(q) with n,q>m, but only with n>m>q. A136112(A135771(k)) are the corresponding indices of the pentagonal numbers.

Examples

			The first terms of this sequence correspond to the following elements of A136113:
P_5 = P_7 - P_5,
P_23 = P_24 - P_7,
P_51 = P_66 - P_42,
P_71 = P_74 - P_21,
P_72 = P_80 - P_35,
P_99 = P_104 - P_32,
P_123 = P_144 - P_75,
P_239 = P_249 - P_70,
P_263 = P_274 - P_77,
P_311 = P_324 - P_91,
P_359 = P_374 - P_10.
		

Crossrefs

Programs

  • PARI
    P(n)=n*(3*n-1)/2
    isPent(t)=P(sqrtint((t*2)\3)+1)==t
    {for( i=1,999, for( j=1,i-1, isPent(P(i)+P(j))|next; for( k=i+1,(P(i)-1)\3, isPent(P(i)+P(k))&next(3)); print1(i", "); next(2)))}

Formula

Equals the difference set A136112 \ A135768.

A136118 Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number.

Original entry on oeis.org

5, 4, 7, 12, 19, 17, 25, 20, 10, 28, 45, 42, 39, 17, 37, 21, 36, 35, 13, 33, 65, 28, 67, 32, 52, 40, 74, 31, 70, 85, 35, 16, 60, 70, 77, 68, 42, 30, 105, 76, 59, 26, 74, 49, 115, 19, 125, 115, 102, 110, 92, 56, 103, 29, 145, 100, 114, 77, 92, 47, 63, 108, 152, 95, 22, 116
Offset: 1

Views

Author

M. F. Hasler, Dec 25 2007

Keywords

Examples

			a(1)=5 is the least integer m>0 such that A136117(1)-P(m) is a pentagonal number, namely P(7)-P(5)=70-35=35=P(5).
a(2)=4 is the least integer m>0 such that A136117(2)-P(m) is a pentagonal number, namely P(8)-P(4)=92-22=70=P(7).
		

Crossrefs

Programs

  • PARI
    A136118vect(n,i=-1)=vector(n,k,until(0,for(j=2,#n=sum2sqr((i+=6)^2+1),n[j]%6==[5,5]||next;n=n[j];break(2)));n[1]\6+1) /* This uses sum2sqr(), cf. A133388. Below some simpler but much slower code. */
    my(P=A000326(n)=n*(3*n-1)/2,isPent(t)=P(sqrtint(t*2\3)+1)==t); for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j))&print1(j",")||next(2)))

A135768 Indices of pentagonal numbers > 0 which are not the difference of 2 other pentagonal numbers > 0.

Original entry on oeis.org

1, 2, 3, 6, 8, 9, 11, 15, 18, 24, 27, 54, 81, 96, 128, 135, 162, 216, 243, 288, 303, 384, 423, 459, 486, 519, 591, 639, 648, 683, 729, 783, 864, 879, 891, 1215, 1458, 1719, 1944, 2031, 2043, 2048, 2151, 2187, 2463, 2799, 3231, 3456, 3543, 3879, 3903, 4023
Offset: 1

Views

Author

R. J. Mathar and M. F. Hasler, Feb 07 2008

Keywords

Comments

A subsequence of A136112, obtained by omitting A136112(A135771(k)), k=1,2,3,... ; i.e. those which are not the difference of two larger pentagonal numbers, but the difference of a larger and a smaller (or equal) pentagonal number. Sequence A135769 has the pentagonal numbers corresponding to these indices.

Examples

			Indices of the following numbers are not here but in A136112:
P_5 = P_7 - P_5
P_23 = P_24 - P_7
P_51 = P_66 - P_42
P_71 = P_74 - P_21
P_72 = P_80 - P_35
P_99 = P_104 - P_32
P_123 = P_144 - P_75
P_239 = P_249 - P_70
P_263 = P_274 - P_77
P_311 = P_324 - P_91
P_359 = P_374 - P_105
		

Crossrefs

Programs

  • Mathematica
    Select[Range[100], Reduce[# (3 # - 1) == x (3 x - 1) - y (3 y - 1) && x > 0 && y > 0, {x, y}, Integers] == False &] (* T. D. Noe, Dec 05 2011 *)
  • PARI
    P(n)=n*(3*n-1)/2
    isPent(t)=P(sqrtint((t*2)\3)+1)==t
    for( i=1,999,for( j=1,(P(i)-1)\3, isPent(P(i)+P(j))&next(2)); print1(i","))

Formula

P(n)=n*(3*n-1)/2 <=> n*(n-1/3) = (2/3)*P(n), thus m = P(n) <=> m = P([sqrt(2m/3)]+1) and m = P(n) <=> 24m+1 = (6n-1)^2, useful for investigating the possibility of write P(n)=P(n')+P(n"): this is possible whenever (6n-1)^2 = (6n'-1)^2 + (6n"-1)^2.

Extensions

Extended by T. D. Noe, Dec 05 2011

A136113 Pentagonal numbers > 0 which are not the difference of two larger pentagonal numbers.

Original entry on oeis.org

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
Offset: 1

Views

Author

M. F. Hasler, Dec 15 2007, Feb 07 2008

Keywords

Examples

			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.
		

Crossrefs

Programs

  • 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)","))

Formula

a(n)=A000326(A136112(n)). A number m is in this sequence iff A136114(m) = 0 iff A136115(m) = 0.

Extensions

a(34)-a(39) from Donovan Johnson, Sep 05 2008

A136117 Pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

Original entry on oeis.org

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
Offset: 1

Views

Author

M. F. Hasler, Dec 15 2007; corrected Dec 25 2007

Keywords

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

Examples

			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).
		

Crossrefs

Programs

  • 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 */
    A136117(n,i)=until(!n--,i=A136117next(i));i \\ M. F. Hasler, Dec 25 2007

Formula

a(n) = A000326(A136116(n)) = A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.

A135769 Pentagonal numbers > 0 which are not the difference of two other pentagonal numbers > 0.

Original entry on oeis.org

1, 5, 12, 51, 92, 117, 176, 330, 477, 852, 1080, 4347, 9801, 13776, 24512, 27270, 39285, 69876, 88452, 124272, 137562, 220992, 268182, 315792, 354051, 403782, 523626, 612162, 629532, 699392, 796797, 919242, 1119312, 1158522, 1190376
Offset: 1

Views

Author

R. J. Mathar and M. F. Hasler, Feb 07 2008

Keywords

Comments

A subsequence of A136113, obtained by omitting A136113(A135771(k)), k=1,2,3,... ; i.e. those which are not the difference of two larger pentagonal numbers, but the difference of a larger and a smaller pentagonal number.
The definition ("...two other...") excludes the case P(n) = P(m)-P(n), cf. comment by R. J. Mathar in A000326.

Examples

			See A135768 for a list of P(n) which are in A136113 but not in A135769.
		

Crossrefs

Programs

  • PARI
    P(n)=n*(3*n-1)/2
    isPent(t)=P(sqrtint((t*2)\3)+1)==t
    for( i=1,999,for( j=1,(P(i)-1)\3, isPent(P(i)+P(j))&j!=i&next(2)); print1(P(i)","))

Formula

P(n)=n*(3*n-1)/2 <=> n*(n-1/3) = (2/3)*P(n), thus m = P(n) <=> m = P([sqrt(2m/3)]+1)
and m = P(n) <=> 24m+1 = (6n-1)^2, useful for investigating the possibility of writing P(n)=P(n')+P(n"): this is possible whenever (6n-1)^2=(6n'-1)^2+(6n"-1)^2-1.

A136115 Index m of least pentagonal number P(m) > P(n) such that P(m)+P(n) is again a pentagonal number, 0 if no such m exists.

Original entry on oeis.org

0, 0, 0, 7, 0, 0, 23, 0, 0, 48, 0, 22, 82, 47, 0, 125, 26, 0, 22, 37, 71, 238, 0, 0, 26, 166, 0, 52, 207, 147, 117, 99, 87, 572, 72, 67, 63, 357, 57, 110, 416, 51, 917, 82, 47, 1050, 217, 380, 167, 246, 0, 97, 697, 0, 374, 191, 537, 1672, 152, 112, 136, 380, 215, 2037, 68
Offset: 1

Views

Author

M. F. Hasler, Dec 15 2007

Keywords

Examples

			a(1..3)=0 since P(1),P(2),P(3) cannot be written as difference of 2 other pentagonal numbers > 0.
a(4)=7 since P(7)=70 is the least pentagonal number > P(4)=22 such that their sum is again a pentagonal number, P(8).
		

Crossrefs

Programs

  • 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(j",")|next(2));print1(0","))

Formula

a(n)=0 iff n is in A136112 iff A000326(n) is in A136113.

A136114 Least pentagonal number P(m) > P(n) such that P(m)+P(n) is again a pentagonal number, 0 if no such m exists.

Original entry on oeis.org

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
Offset: 1

Views

Author

M. F. Hasler, Dec 15 2007

Keywords

Examples

			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).
		

Crossrefs

Programs

  • 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","))

Formula

a(n)=A000326(A136115(n)). a(n)=0 iff n is in A136112 iff A000326(n) is in A136113.

A136116 Indices of pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

Original entry on oeis.org

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
Offset: 1

Views

Author

M. F. Hasler, Dec 15 2007; corrected Dec 25 2007

Keywords

Examples

			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).
		

Crossrefs

Programs

  • 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))
    A136116(n,i=0)=until(!n--,i=A136116next(i));i \\ M. F. Hasler, Dec 25 2007

Formula

A000326(a(n))=A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.

A137694 Numbers k such that 6k^2-2k = 3n^2-n for some integer n>0.

Original entry on oeis.org

5, 5577, 6435661, 7426747025, 8570459630997, 9890302987423321, 11413401077026881245, 13171054952586033533217, 15199386001883205670450981, 17540078275118266757666898665, 20241235130100477955141930608237, 23358367800057676441967030255006641
Offset: 1

Views

Author

M. F. Hasler, Feb 08 2008

Keywords

Comments

Also indices of pentagonal numbers which are half of some other pentagonal number: see A137693 for more details, comments and links.

Crossrefs

Programs

  • PARI
    vector(20,i, (v=if(i>1,[577,408;816,577]*v-[164;232], [5;7]))[1,1])

Formula

a(n) = f^{2n-2}(5,7)[1], where f(x,y) = (577x + 408y - 164, 816x + 577y - 232).
a(n) = (5,7,1,5,7,1,...) (mod 10).
G.f.: -x*(5-198*x+x^2) / ( (x-1)*(x^2-1154*x+1) ). - R. J. Mathar, Apr 17 2011
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