A016135 -id:A016135 - cp's OEIS frontend

A016325 Expansion of 1/((1-2x)(1-10x)(1-11x)).

1, 23, 377, 5395, 71841, 915243, 11317657, 136994195, 1631936081, 19201296763, 223714264137, 2585856904995, 29694425953921, 339138685491083, 3855525540397817, 43660780944367795, 492768590388029361

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (23,-152,220)

Cf. A016134, A016135.

[(2*11^(n+2) +2^n-225*10^n)/18 : n in [0..20]]; // Vincenzo Librandi, Oct 09 2011

CoefficientList[Series[1/((1 - 2 x) (1 - 10 x) (1 - 11 x)), {x, 0, 16}], x] (* Michael De Vlieger, Jan 31 2018 *)

Vec(1/((1-2*x)*(1-10*x)*(1-11*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

[(11^n - 2^n)/9-(10^n - 2^n)/8 for n in range(2,19)] # Zerinvary Lajos, Jun 05 2009

From Zerinvary Lajos, Jun 05 2009 [corrected by R. J. Mathar, Mar 14 2011]: (Start)
a(n) = 11^(n+2)/9 + 2^(n-1)/9 - 25*10^n/2.
a(n) = A016135(n+1) - A016134(n+1). (End)
a(n) = 21*a(n-1) - 110*a(n-2) + 2^n. - Vincenzo Librandi, Oct 09 2011

A211072 Sum of numbers with no '0' decimal digits whose sum of digits equals n.

Original entry on oeis.org

0, 1, 13, 147, 1625, 17891, 196833, 2165227, 23817625, 261994131, 2881935943, 31701296375, 348714262017, 3835856884757, 42194425724149, 464138682802857, 5105525508895321, 56160780576260645, 617768586100819485, 6795454444489330049, 74749998860563784655

Offset: 0

Laurent Desnogues and Charles R Greathouse IV, Apr 02 2012

Different from A016135.

			2 and 11 are the only numbers without 0's which have digit sum 2, so a(2) = 2 + 11 = 13.

Alois P. Heinz, Table of n, a(n) for n = 0..961 (terms n = 1..31 from Laurent Desnogues)
Project Euler, Problem 377: Sum of digits, experience 13
Tadao Takaoku, A two-level algorithm for generating multiset permutations, RIMS Kokyuroku 1644 (2009), pp. 95-109.
Index entries for linear recurrences with constant coefficients, signature (11,1,-9,-19,-29,-39,-49,-59,-69,-90,-80,-70,-60,-50,-40,-30,-20,-10).

Cf. A007953, A016135, A052382, A130835, A258800.

b:= proc(n) option remember; `if`(n=0, [1, 0], add((p->
      [p[1], p[2]*10+p[1]*d])(b(n-d)), d=1..min(n, 9)))
    end:
a:= n-> b(n)[2]:
seq(a(n), n=0..23);  # Alois P. Heinz, Feb 19 2020

G.f.: x*(9*x^8 + 8*x^7 + 7*x^6 + 6*x^5 + 5*x^4 + 4*x^3 + 3*x^2 + 2*x + 1)/((x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x - 1)*(10*x^9 + 10*x^8 + 10*x^7 + 10*x^6 + 10*x^5 + 10*x^4 + 10*x^3 + 10*x^2 + 10*x - 1)). - Yurii Ivanov, Jul 06 2021

a(0)=0 prepended by Alois P. Heinz, Feb 19 2020

A255242 Calculate the aliquot parts of a number n and take their sum. Then repeat the process calculating the aliquot parts of all the previous aliquot parts and add their sum to the previous one. Repeat the process until the sum to be added is zero. Sequence lists these sums.

Original entry on oeis.org

0, 1, 1, 4, 1, 8, 1, 12, 5, 10, 1, 30, 1, 12, 11, 32, 1, 36, 1, 38, 13, 16, 1, 92, 7, 18, 19, 46, 1, 74, 1, 80, 17, 22, 15, 140, 1, 24, 19, 116, 1, 90, 1, 62, 51, 28, 1, 256, 9, 62, 23, 70, 1, 136, 19, 140, 25, 34, 1, 286, 1, 36, 61, 192, 21, 122, 1, 86, 29, 114

Offset: 1

Paolo P. Lava, Feb 19 2015

nonn

a(n) = 1 if n is prime.

			The aliquot parts of 8 are 1, 2, 4 and their sum is 7.
Now, let us calculate the aliquot parts of 1, 2 and 4:
1 => 0;  2 => 1;  4 => 1, 2.  Their sum is 0 + 1 + 1 + 2 = 4.
Let us calculate the aliquot parts of 1, 1, 2:
1 => 0;  1 = > 0; 2 => 1. Their sum is 1.
We have left 1: 1 => 0.
Finally, 7 + 4 + 1 = 12. Therefore a(8) = 12.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Paolo P. Lava)
Jon Maiga, Computer-generated formulas for A255242, Sequence Machine.

Cf. A001047, A001065, A001787, A006516, A016127, A016130, A016135, A255243, A330575.

Sequences that appear in the convolution formulas: A000203, A001065, A051953, A062790, A067824, A074206, A157658, A174725, A174726, A191161, A211779, A245211, A323910, A253249.

with(numtheory): P:=proc(q) local a,b,c,k,n,t,v;
for n from 1 to q do b:=0; a:=sort([op(divisors(n))]); t:=nops(a)-1;
while add(a[k],k=1..t)>0 do b:=b+add(a[k],k=1..t); v:=[];
for k from 2 to t do c:=sort([op(divisors(a[k]))]); v:=[op(v),op(c[1..nops(c)-1])]; od;
a:=v; t:=nops(a); od; print(b); od; end: P(10^3);

f[s_] := Flatten[Most[Divisors[#]] & /@ s]; a[n_] := Total@Flatten[FixedPointList[ f, {n}]] - n; Array[a, 100] (* Amiram Eldar, Apr 06 2019 *)

ali(n) = setminus(divisors(n), Set(n));
a(n) = my(list = List(), v = [n]); while (#v, my(w = []); for (i=1, #v, my(s=ali(v[i])); for (j=1, #s, w = concat(w, s[j]); listput(list, s[j]));); v = w;); vecsum(Vec(list)); \\ Michel Marcus, Jul 15 2023

a(1) = 0.
a(2^k) = k*2^(k-1) = A001787(k), for k>=1.
a(n^k) = (n^k-2^k)/(n-2), for n odd prime and k>=1.
In particular:
a(3^k)  = A001047(k-1);
a(5^k)  = A016127(k-1);
a(7^k)  = A016130(k-1);
a(11^k) = A016135(k-1).
From Antti Karttunen, Nov 22 2024: (Start)
a(n) = A330575(n) - n.
Also, following formulas were conjectured by Sequence Machine:
a(n) = (A191161(n)-n)/2.
a(n) = Sum_{d|n} A001065(d)*A074206(n/d). [Compare to David A. Corneth's Apr 13 2020 formula for A330575]
a(n) = Sum_{d|n} A051953(d)*A067824(n/d).
a(n) = Sum_{d|n} A000203(d)*A174726(n/d).
a(n) = Sum_{d|n} A062790(d)*A253249(n/d).
a(n) = Sum_{d|n} A157658(d)*A191161(n/d).
a(n) = Sum_{d|n} A174725(d)*A211779(n/d).
a(n) = Sum_{d|n} A245211(d)*A323910(n/d).
(End)

A139740 a(n) = 11^n - 2^n.

Original entry on oeis.org

0, 9, 117, 1323, 14625, 161019, 1771497, 19487043, 214358625, 2357947179, 25937423577, 285311668563, 3138428372625, 34522712135739, 379749833566857, 4177248169382883, 45949729863506625, 505447028499162699, 5559917313491969337, 61159090448414022003, 672749994932558960625

Offset: 0

N. J. A. Sloane, May 20 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (13,-22).

Cf. A000079, A001020, A016135.

[11^n-2^n: n in [0..30]]; // Vincenzo Librandi, Jun 02 2011

Table[11^n - 2^n, {n, 0, 20}] (* or *)
LinearRecurrence[{13, -22}, {0, 9}, 21] (* Paolo Xausa, Mar 16 2025 *)

a(n)=11^n-2^n \\ Charles R Greathouse IV, Mar 26 2012

[11**n-2**n for n in range(20)] # Gennady Eremin, Mar 05 2022

[11^n - 2^n for n in range(0,20)] # Zerinvary Lajos, Jun 05 2009

a(n) = 13*a(n-1) - 22*a(n-2). - Vincenzo Librandi, Jun 02 2011
a(n) = A001020(n) - A000079(n). - Michel Marcus, Feb 28 2022
a(n) = 9*A016135(n-1), n > 0. - Bernard Schott, Mar 09 2022
E.g.f.: exp(2*x)*(exp(9*x) - 1). - Stefano Spezia, Mar 09 2025
G.f.: 9*x/((1-2*x)*(1-11*x)). - Elmo R. Oliveira, Mar 15 2025

A016633 Expansion of g.f. 1/((1-2x)(1-11x)(1-12*x)).

Original entry on oeis.org

1, 25, 447, 6989, 101759, 1417941, 19180519, 253983853, 3309800367, 42599540357, 542895780791, 6863463633117, 86197420501375, 1076563471968373, 13382900349107463, 165700329729679181, 2044564737700501583, 25152545442794015589, 308625999807796411735, 3778261997130507936445

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (25,-178,264).

Cf. A006516, A016127, A016129, A016130, A016311, A016316, A016321, A016325. - Zerinvary Lajos, Jun 05 2009

Cf. A016135, A016136.

[(648*12^n +2^(n+1)-5*11^(n+2))/45 : n in [0..20]]; // Vincenzo Librandi, Oct 09 2011

CoefficientList[Series[1/((1 - 2 x) (1 - 11 x) (1 - 12 x)), {x, 0, 15}], x] (* Michael De Vlieger, Jan 31 2018 *)

Vec(1/((1-2*x)*(1-11*x)*(1-12*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

[(12^n - 2^n)/10-(11^n - 2^n)/9 for n in range(2,18)] # Zerinvary Lajos, Jun 05 2009

From Vincenzo Librandi, Oct 09 2011: (Start)
a(n) = (648*12^n + 2^(n+1) - 5*11^(n+2))/45.
a(n) = 23*a(n-1) - 132*a(n-2) + 2^n.
a(n) = 25*a(n-1) - 178*a(n-2) + 264*a(n-3), n >= 3. (End)
From Elmo R. Oliveira, Mar 26 2025: (Start)
E.g.f.: exp(2*x)*(648*exp(10*x) - 605*exp(9*x) + 2)/45.
a(n) = A016136(n+1) - A016135(n+1). (End)

A096044 Triangle read by rows: T(n,k) = (n+1,k)-th element of (M^10-M)/9, where M is the infinite lower Pascal's triangle matrix, 1<=k<=n.

Original entry on oeis.org

1, 11, 2, 111, 33, 3, 1111, 444, 66, 4, 11111, 5555, 1110, 110, 5, 111111, 66666, 16665, 2220, 165, 6, 1111111, 777777, 233331, 38885, 3885, 231, 7, 11111111, 8888888, 3111108, 622216, 77770, 6216, 308, 8, 111111111, 99999999, 39999996, 9333324, 1399986, 139986, 9324, 396, 9

Offset: 1

Gary W. Adamson, Jun 17 2004

			Triangle T(n,k) begins:
       1;
      11,     2;
     111,    33,     3;
    1111,   444,    66,    4;
   11111,  5555,  1110,  110,   5;
  111111, 66666, 16665, 2220, 165, 6;
  ...

Alois P. Heinz, Rows n = 1..141, flattened

Cf. A007318. First column gives A000042. Row sums give A016135.

Cf. A096034, A096035, A096039, A096040, A096041, A096042, A096043.

P:= proc(n) option remember; local M; M:= Matrix(n, (i, j)-> binomial(i-1, j-1)); (M^10-M)/9 end: T:= (n, k)-> P(n+1)[n+1, k]: seq(seq(T(n, k), k=1..n), n=1..11);  # Alois P. Heinz, Oct 07 2009

P[n_] := P[n] = With[{M = Array[Binomial[#1-1, #2-1]&, {n, n}]}, (MatrixPower[M, 10] - M)/9]; T[n_, k_] := P[n+1][[n+1, k]]; Table[ Table[T[n, k], {k, 1, n}], {n, 1, 11}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Alois P. Heinz *)

Edited and more terms from Alois P. Heinz, Oct 07 2009

cp's OEIS Frontend

A016325 Expansion of 1/((1-2x)(1-10x)(1-11x)).

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A211072 Sum of numbers with no '0' decimal digits whose sum of digits equals n.

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A255242 Calculate the aliquot parts of a number n and take their sum. Then repeat the process calculating the aliquot parts of all the previous aliquot parts and add their sum to the previous one. Repeat the process until the sum to be added is zero. Sequence lists these sums.

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A139740 a(n) = 11^n - 2^n.

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A016633 Expansion of g.f. 1/((1-2*x)*(1-11*x)*(1-12*x)).

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A096044 Triangle read by rows: T(n,k) = (n+1,k)-th element of (M^10-M)/9, where M is the infinite lower Pascal's triangle matrix, 1<=k<=n.

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A016633 Expansion of g.f. 1/((1-2x)(1-11x)(1-12*x)).