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 1-8 of 8 results.

A013959 a(n) = sigma_11(n), the sum of the 11th powers of the divisors of n.

Original entry on oeis.org

1, 2049, 177148, 4196353, 48828126, 362976252, 1977326744, 8594130945, 31381236757, 100048830174, 285311670612, 743375541244, 1792160394038, 4051542498456, 8649804864648, 17600780175361, 34271896307634, 64300154115093, 116490258898220, 204900053024478
Offset: 1

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Author

Keywords

Comments

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
Related to congruence properties of the Ramanujan tau function since A000594(n) == a(n) (mod 691) = A046694(n). - Benoit Cloitre, Aug 28 2002

Crossrefs

Programs

Formula

G.f.: Sum_{k>=1} k^11*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
Dirichlet g.f.: zeta(s-11)*zeta(s). - Ilya Gutkovskiy, Sep 10 2016
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(11*e+11)-1)/(p^11-1).
Sum_{k=1..n} a(k) = zeta(12) * n^12 / 12 + O(n^13). (End)

A281788 a(n) = (A013967(n) - A037945(n))/174611.

Original entry on oeis.org

0, 3, 6656, 1574235, 109234176, 3489819540, 65281655808, 825351571995, 7736349470720, 57270269768634, 350259092774400, 1829670576438068, 8372440970643456, 34226453991167880, 126958657929489408, 432721923827171355, 1369171676955783168, 4056082931864408991, 11330441127202890240, 30026115193307387658, 75874353000273633280, 183636989491548765276
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2017

Keywords

Examples

			a(1) = (1 - 1)/174611 = 0.
a(2) = (524289 - 456)/174611 = 3.
a(3) = (1162261468 - 50652)/174611 = 6656.
		

Crossrefs

A281876 a(n) = (A013963(n) - A027364(n))/3617.

Original entry on oeis.org

0, 9, 3968, 296865, 8437248, 129997260, 1312568064, 9727799265, 56923182080, 276480648702, 1154893046400, 4259743681004, 14151477247488, 43011568291320, 121065502097664, 318760489739745, 791380439553024, 1865315725321293, 4197159808767360, 9059718006875214
Offset: 1

Views

Author

Seiichi Manyama, Feb 01 2017

Keywords

Examples

			a(1) = (1 - 1)/3617 = 0.
a(2) = (32769 - 216)/3617 = 9.
a(3) = (14348908 - (-3348))/3617 = 3968.
		

Crossrefs

A281928 a(n) = (A013965(n) - A037944(n))/43867.

Original entry on oeis.org

0, 3, 2944, 391635, 17392128, 385866060, 5303086848, 51332824275, 380176030720, 2279635315794, 11522261136000, 50576242992268, 197196432781824, 695091512105880, 2246019242126592, 6728295917456595, 18857917384178688, 49830812542200039
Offset: 1

Views

Author

Seiichi Manyama, Feb 02 2017

Keywords

Examples

			a(1) = (1 - 1)/43867 = 0.
a(2) = (131073 - (-528))/43867 = 3.
a(3) = (129140164 - (-4284))/43867 = 2944.
		

Crossrefs

A281956 a(n) = (A013969(n) - A037946(n))/77683.

Original entry on oeis.org

0, 27, 134656, 56615355, 6138243072, 282390755580, 7190065585152, 118730950577595, 1408531971420160, 12872835457479666, 95262154452748800, 592216338844654972, 3180419513581234176, 15078667591360144440, 64208193499209765888, 248996850497620053435
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2017

Keywords

Examples

			a(1) = (1 - 1)/77683 = 0.
a(2) = (2097153 - (-288))/77683 = 27.
a(3) = (10460353204 - (-128844))/77683 = 134656.
		

Crossrefs

A279889 a(n) = Sum_{k=1..n-1} sigma_5(k)*sigma_5(n-k).

Original entry on oeis.org

0, 1, 66, 1577, 18218, 135550, 738236, 3207785, 11714718, 37347144, 106499470, 277489886, 668981686, 1512360404, 3228797252, 6570019945, 12793050456, 24001960051, 43483452090, 76485144056, 130752372320, 218220937122, 355664809556, 568293832670, 889969136158
Offset: 1

Views

Author

Seiichi Manyama, Dec 22 2016

Keywords

Comments

In 1916, Ramanujan found the following identity. tau(n) = sigma_11(n) - 691/756 * (sigma_11(n) - sigma_5(n) + 252 * a(n)). This implies tau(n) == sigma_11(n) mod 691.

References

  • Tom M. Apostol, Modular Functions and Dirichlet Series in Number Theory,Springer-Verlag, 1976. See p. 140, exercise 10.
  • Srinivasa Ramanujan, Collected papers, ed. G. H. Hardy et al., Cambridge, 1927, pp. 136-162.

Crossrefs

Cf. Sum_{k=1..n-1} sigma_m(k)*sigma_m(n-k): A087115 (m=3), this sequence (m=5).

Programs

  • Mathematica
    a[n_] := (65 * DivisorSigma[11, n] + 691 * DivisorSigma[5, n] - 756 * RamanujanTau[n]) / 174132; Array[a, 25] (* Amiram Eldar, Jan 07 2025 *)
  • PARI
    a(n) = sum(k=1, n-1, sigma(k, 5)*sigma(n-k, 5)) \\ Felix Fröhlich, Jan 01 2017
    
  • PARI
    a(n) = {my(f = factor(n)); (65 * sigma(f, 11) + 691 * sigma(f, 5) - 756 * ramanujantau(n)) / 174132;} \\ Amiram Eldar, Jan 07 2025

Formula

A027860(n) = (sigma_11(n) - sigma_5(n) + 252*a(n))/756.

A281979 a(n) = (A281959(n) - A037947(n))/657931.

Original entry on oeis.org

0, 51, 1287808, 1711273635, 452970333696, 43211657266860, 2038311950075136, 57420813107839395, 1091144797392901120, 15199162675148592018, 164678453263146595200, 1449942615368630353516, 10725152052216567264768, 68394401763888606334680
Offset: 1

Views

Author

Seiichi Manyama, Feb 04 2017

Keywords

Examples

			a(1) = (1 - 1)/657931 = 0.
a(2) = (33554433 - (-48))/657931 = 51.
a(3) = (847288609444 - (-195804))/657931 = 1287808.
		

Crossrefs

A337032 a(n) = (n*sigma_9(n) - tau(n))/7 = (A282254(n) - A000594(n))/7, where tau is Ramanujan's tau, sigma_9(n) = Sum_{d divides n} d^9.

Original entry on oeis.org

0, 150, 8400, 150300, 1394400, 8656200, 40356000, 153679800, 498153600, 1431378900, 3705270000, 8863150800, 19694152800, 41402744400, 82382680800, 157380332400, 288000115200, 511088547150, 875865085200, 1465721632200, 2382961862400, 3801687211800, 5918070367200, 9075809181600
Offset: 1

Views

Author

Jianing Song, Aug 12 2020

Keywords

Comments

D. H. Lehmer shows that tau(n) == n*sigma_9(n) (mod 7), so a(n) is an integer for all n. Furthermore, if n == 3, 5, 6 (mod 7) then tau(n) == n*sigma_9(n) (mod 49). See the Wikipedia link below. It seems that the latter congruence also holds for most of the other numbers. Among the 571 numbers in [1, 1000] congruent to 0, 1, 2, 4 modulo 7, tau(n) == n*sigma_9(n) holds for 311 n's, and among the 5715 numbers in [1, 10000] congruent to 0, 1, 2, 4 modulo 7, the congruence holds for 3492 n's.
It seems that 150 divides a(n) for all n. There are no counterexamples for n <= 10000.
Number of n's in [2, N] which satisfy the higher-order congruence tau(n) == n*sigma_9(n) (mod 7^e) but not tau(n) == n*sigma_9(n) (mod 7^(e+1)):
N = 1000:
e | n == 3, 5, 6 (mod 7) | n == 0, 1, 2, 4 (mod 7) | total
---+----------------------+-------------------------+-------
1 | 0 | 260 | 260
---+----------------------+-------------------------+-------
2 | 358 | 80 | 438
---+----------------------+-------------------------+-------
3 | 45 | 195 | 240
---+----------------------+-------------------------+-------
4 | 24 | 28 | 52
---+----------------------+-------------------------+-------
5 | 2 | 5 | 7
---+----------------------+-------------------------+-------
6 | 0 | 2* | 2
* n = 686, 942.
N = 10000:
e | n == 3, 5, 6 (mod 7) | n == 0, 1, 2, 4 (mod 7) | total
---+----------------------+-------------------------+-------
1 | 0 | 2223 | 2223
---+----------------------+-------------------------+-------
2 | 3368 | 728 | 4096
---+----------------------+-------------------------+-------
3 | 466 | 2280 | 2746
---+----------------------+-------------------------+-------
4 | 397 | 384 | 781
---+----------------------+-------------------------+-------
5 | 46 | 87 | 133
---+----------------------+-------------------------+-------
6 | 6 | 12 | 18
---+----------------------+-------------------------+-------
7 | 2** | 0 | 2
** n = 5185, 9021.

Examples

			a(2) = (n*sigma_9(2) - tau(2))/7 = (2*(1^9+2^9) - (-24))/7 = 1050/7 = 150;
a(3) = (n*sigma_9(3) - tau(3))/7 = (3*(1^9+3^9) - 252)/7 = 58800/7 = 8400.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := (n * DivisorSigma[9, n] - RamanujanTau[n]) / 7; Array[a, 24] (* Amiram Eldar, Jan 10 2025 *)
  • PARI
    a(n) = (n*sigma(n, 9) - polcoeff( x * eta(x + x * O(x^n))^24, n))/7;
Showing 1-8 of 8 results.