A049000 -id:A049000 - cp's OEIS frontend

A072692 Sum of sigma(j) for 1<=j<=10^n, where sigma(j) is the sum of the divisors of j.

1, 87, 8299, 823081, 82256014, 8224740835, 822468118437, 82246711794796, 8224670422194237, 822467034112360628, 82246703352400266400, 8224670334323560419029, 822467033425357340138978, 82246703342420509396897774, 8224670334241228180927002517

Offset: 0

Rick L. Shepherd, Jul 02 2002

nonn

			For n=1, the sum of sigma(j) for j<=10 is 1+3+4+7+6+12+8+15+13+18=87, so a(1)=87 (=69+18=A049000(1)+A046915(1)).

P. L. Patodia, Seth Troisi and Hiroaki Yamanouchi, Table of n, a(n) for n = 0..36 (terms a(0)-a(18) by P. L. Patodia and a(19)-a(24) by Seth Troisi)
Leonhard Euler, Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs, 1747, The Euler Archive, (Eneström Index) E175.
P. L. Patodia (pannalal(AT)usa.net), PARI program for A072692 and A024916

Compare with A049000. Note that a(n) = A049000(n) + A046915(n).

Cf. A000203 (sigma(n)), A072691 (Pi^2/12), A049000, A046915, A024916, A025281.

for(m=0,10,print1(sum(n=1,k=10^m,n*(k\n)),",")) \\ Improved by M. F. Hasler, Apr 18 2015

A072692(n)=A024916(10^n) \\ This is very efficient, using efficient code of A024916. - M. F. Hasler, Apr 18 2015

[(i, sum([d*(10**i//d) for d in range(1,10**i+1)])) for i in range(8)] # Seth A. Troisi, Jun 27 2010

from math import isqrt
def A072692(n): return -(s:=isqrt(m:=10**n))**2*(s+1)+sum((q:=m//k)*((k<<1)+q+1) for k in range(1,s+1))>>1 # Chai Wah Wu, Oct 23 2023

Asymptotic formula: a(n) ~ Pi^2/12 * 10^2n. See A072691 for Pi^2/12. Observe that A025281 also contains that constant in its asymptotic formula.

More terms from P L Patodia (pannalal(AT)usa.net), Jan 11 2008, Jun 25 2008

Corrected by N. J. A. Sloane, Jun 08 2008, following suggestions from Don Reble and David W. Wilson

A046915 Sum of divisors of 10^n.

Original entry on oeis.org

1, 18, 217, 2340, 24211, 246078, 2480437, 24902280, 249511591, 2497558338, 24987792457, 249938963820, 2499694822171, 24998474116998, 249992370597277, 2499961853010960, 24999809265103951, 249999046325618058, 2499995231628286897

Offset: 0

Enoch Haga

A072692(n) = A049000(n) + a(n).

a(n) is the number of full-dimensional lattices in Z^(n+1) with volume 10. - Álvar Ibeas, Nov 29 2015

			At 10^1 the factors are 1, 2, 5, 10. The sum of these factors is 18: 1 + 2 + 5 + 10.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (18,-97,180,-100).

Cf. A000203 (sigma(n)), A049000, A072692. 10th row of A160870, shifted.

[1/4*(2^(n+1)-1)*(5^(n+1)-1): n in [0..20]]; // Vincenzo Librandi, Oct 03 2011

Table[DivisorSigma[1, 10^n], {n, 0, 18}] (* Jayanta Basu, Jun 30 2013 *)

Vec(-(10*x^2-1)/((x-1)*(2*x-1)*(5*x-1)*(10*x-1)) + O(x^100)) \\ Colin Barker, Jan 27 2015

a(n) = sigma(10^n); \\ Altug Alkan, Dec 04 2015

a(n) = 1/4*(2^(n+1)-1)*(5^(n+1)-1). E.g., a(1) = 1/4*(2^2-1)*(5^2-1) = 18. - Vladeta Jovovic, Dec 18 2001
a(n) = 18*a(n-1)-97*a(n-2)+180*a(n-3)-100*a(n-4). - Colin Barker, Jan 27 2015
G.f.: -(10*x^2-1) / ((x-1)*(2*x-1)*(5*x-1)*(10*x-1)). - Colin Barker, Jan 27 2015

A049030 Sum of sigma(j) for 1<=j<10^n, where sigma(j) = A048050(j) is the sum of the proper divisors >1 of j (excluding 1 and n).

Original entry on oeis.org

16, 3034, 320243, 32226805, 3224444759, 322465138002, 32246681892518, 3224670122682648, 322467031114802292, 32246703322412473945, 3224670334023621455211, 322467033422357645316809, 32246703342390510922780778, 3224670334240928188556405242

Offset: 1

Enoch Haga and Jud McCranie

			For n = 1, the sum of sigma(j), for j < 10 is 0 + 0 + 0 + 2 + 0 + 5 + 0 + 6 + 3 = 16, so a(1) = 16.

Amiram Eldar, Table of n, a(n) for n = 1..36 (calculated from the b-file at A049000)
Carlos Rivera, Problem 23.- Divisors (V) Aliquot sequences, Sociable Numbers, Prime Puzzles and Problems Connection.

Cf. A001065, A046915, A048050, A048995, A049000.

Cf. A072691 (Pi^2/12).

At a(3) = 320243, for example, take a(3) from A049000: 820741 - 500498 = 320243. Compute 500498 from 999*1000/2 = 499500, split evenly and reverse to 500499 - 1 = 500498. Add a 9 and 0 for each successive term.

a(n) = A049000(n) - 10^n * (10^n + 1) / 2 + 2 ~ (Pi^2/12 - 1/2) * 10^(2*n). - Amiram Eldar, Feb 16 2020

More terms from Amiram Eldar, Feb 16 2020

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A072692 Sum of sigma(j) for 1<=j<=10^n, where sigma(j) is the sum of the divisors of j.

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A046915 Sum of divisors of 10^n.

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A049030 Sum of sigma(j) for 1<=j<10^n, where sigma(j) = A048050(j) is the sum of the proper divisors >1 of j (excluding 1 and n).

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