A002064 -id:A002064 - cp's OEIS frontend

A070940 Number of digits that must be counted from left to right to reach the last 1 in the binary representation of n.

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 4, 2, 4, 3, 4, 1, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 1, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 2, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7, 6, 7, 3, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7

Offset: 1

N. J. A. Sloane, May 18 2002

Length of longest carry sequence when adding numbers <= n to n in binary representation: a(n) = T(n, A080079(n)) and T(n,k) <= a(n) for 1 <= k <= n, with T defined as in A080080. - Reinhard Zumkeller, Jan 26 2003
a(n+1) is the number of distinct values of gcd(2^n, binomial(n,j)) (or, equivalently, A007814(binomial(n,j))) arising for j=0..n-1. Proof using Kummer's Theorem given by Marc Schwartz. - Labos Elemer, Apr 23 2003
E.g., n=10: 10th row of Pascal's triangle = {1,10,45,120,210,252,210,120,45,10,1}, largest powers of 2 dividing binomial coefficients is: {1,2,1,8,2,4,2,8,1,2,1}; including distinct powers of 2, thus a(10)=4. If m=-1+2^k, i.e., m=0,1,3,7,15,31,... then a(m)=1. This corresponds to "odd rows" of Pascal's triangle. - Labos Elemer
Smallest x > 0 for which a(x)=n equals 2^n. - Labos Elemer
a(n) <= A070939(n), a(n) = A070939(n) iff n is odd, where A070939(n) = floor(log_2(n)) + 1. - Reinhard Zumkeller, Jan 26 2003
Can be regarded as a table with row n having 2^(n-1) columns, with odd columns repeating the previous row, and even columns containing the row number. - Franklin T. Adams-Watters, Nov 08 2011
It appears that a(n) is the greatest number in a periodicity equivalence class defined at A269570; e.g., the 5 classes for n = 35 are (1, 1, 2, 2, 6), (1, 1, 1, 1, 4, 2, 2), (3), (1, 3), (1, 2); in these the greatest number is 6, so that a(35) = 6. - Clark Kimberling, Mar 01 2016
Number of binary digits of the largest odd factor of n. - Andres Cicuttin, May 18 2017

			a(10)=3 is the number of digits that must be counted from left to right to reach the last 1 in 1010, the binary representation of 10.
The table starts:
  1
  1 2
  1 3 2 3
  1 4 3 4 2 4 3 4

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A070939, A001511. Differs from A002487 around 11th term.
Cf. A000005, A007318, A000079, A082907, A082908.
Bisections give A070941 and this sequence (again).
Cf. A002064 (row sums), A199570.

a070940 = maximum . a080080_row  -- Reinhard Zumkeller, Apr 22 2013

A070940 := n -> if n mod 2 = 0 then A070939(n)-A001511(n/2) else A070939(n); fi;

Table[Length[Union[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}]]], {n, 0, 256}]
f[n_] := Position[ IntegerDigits[n, 2], 1][[ -1, 1]]; Table[ f[n], {n, 105}] (* Robert G. Wilson v, Dec 01 2004 *)
(* By exploiting the "positional" regularity of the sequence *)
b = {}; a = {1, 1};
Do[a = Riffle[a, j];
  b = AppendTo[b, a[[1 ;; Floor[Length[a]/2]]]] // Flatten, {j, 1, 10}];
Print[b[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *)
(* By following the alternative definition "Number of binary digits of the largest integer odd factor of n" *)
c = Table[IntegerDigits[n/(2^IntegerExponent[n, 2]), 2] // Length , {n,
    2^10 - 1}];
Print[c[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *)
lidn[n_]:=Module[{idn=IntegerDigits[n,2]},idn=If[Last[idn]==0,Flatten[ Most[ Split[ idn]]],idn];Length[idn]]; Array[lidn,100] (* Harvey P. Dale, Oct 18 2020 *)
Table[IntegerLength[FromDigits[Reverse[IntegerDigits[n,2]]]],{n,100}] (* Harvey P. Dale, Jan 26 2025 *)

def A070940(n):
    while n%2 == 0:
        n = n//2
    a = 0
    while n != 0:
        n, a = n//2, a+1
    return a
n = 0
while n < 100:
    n = n+1
    print(n,A070940(n)) # A.H.M. Smeets, Aug 19 2019

def A070940(n): return n.bit_length()-(~n&n-1).bit_length() # Chai Wah Wu, Jul 13 2022

blocklevel <- 7  # by choice
a <- 1
for(m in 0:blocklevel)
  for(k in 0:(2^m-1)){
    a[2^(m+1)+2*k  ] <-  a[2^m+k]
    a[2^(m+1)+2*k+1] <-  m + 2
}
a
# Yosu Yurramendi, Aug 08 2019

a(n) = floor(log_2(n)) - A007814(n) = A070939(n) - A007814(n).
a(n) = f(n, 1), f(n, k) = if n=1 then k else f(floor(n/2), k+(if k>1 then 1 else n mod 2)). - Reinhard Zumkeller, Feb 01 2003
G.f.: Sum_{k>=0} (t/(1-t^2)) * (1 + Sum_{L>=1} t^2^L), where t=x^2^k. - Ralf Stephan, Mar 15 2004
a(n) = A070939(A000265(n)). - Andres Cicuttin, May 19 2017
a(1) = 1 and for m >= 0, 0 <= k < 2^m, a(2^(m+1)+2*k) = a(2^m+k), a(2^(m+1)+2*k+1) = m+2. - Yosu Yurramendi, Aug 08 2019

Entry revised by Ralf Stephan, Nov 29 2004

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A343656 Array read by antidiagonals where A(n,k) is the number of divisors of n^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 3, 3, 1, 1, 5, 4, 5, 2, 1, 1, 6, 5, 7, 3, 4, 1, 1, 7, 6, 9, 4, 9, 2, 1, 1, 8, 7, 11, 5, 16, 3, 4, 1, 1, 9, 8, 13, 6, 25, 4, 7, 3, 1, 1, 10, 9, 15, 7, 36, 5, 10, 5, 4, 1, 1, 11, 10, 17, 8, 49, 6, 13, 7, 9, 2, 1, 1, 12, 11, 19, 9, 64, 7, 16, 9, 16, 3, 6, 1

Offset: 1

Gus Wiseman, Apr 28 2021

First differs from A343658 at A(4,2) = 5, A343658(4,2) = 6.

As a triangle, T(n,k) = number of divisors of k^(n-k).

			Array begins:
       k=0 k=1 k=2 k=3 k=4 k=5 k=6 k=7
  n=1:  1   1   1   1   1   1   1   1
  n=2:  1   2   3   4   5   6   7   8
  n=3:  1   2   3   4   5   6   7   8
  n=4:  1   3   5   7   9  11  13  15
  n=5:  1   2   3   4   5   6   7   8
  n=6:  1   4   9  16  25  36  49  64
  n=7:  1   2   3   4   5   6   7   8
  n=8:  1   4   7  10  13  16  19  22
  n=9:  1   3   5   7   9  11  13  15
Triangle begins:
  1
  1  1
  1  2  1
  1  3  2  1
  1  4  3  3  1
  1  5  4  5  2  1
  1  6  5  7  3  4  1
  1  7  6  9  4  9  2  1
  1  8  7 11  5 16  3  4  1
  1  9  8 13  6 25  4  7  3  1
  1 10  9 15  7 36  5 10  5  4  1
  1 11 10 17  8 49  6 13  7  9  2  1
  1 12 11 19  9 64  7 16  9 16  3  6  1
  1 13 12 21 10 81  8 19 11 25  4 15  2  1
For example, row n = 8 counts the following divisors:
  1  64  243  256  125  36  7  1
     32  81   128  25   18  1
     16  27   64   5    12
     8   9    32   1    9
     4   3    16        6
     2   1    8         4
     1        4         3
              2         2
              1         1

Seiichi Manyama, Antidiagonals n = 1..140, flattened

Columns k=1..9 of the array give A000005, A048691, A048785, A344327, A344328, A344329, A343526, A344335, A344336.
Row n = 6 of the array is A000290.
Diagonal n = k of the array is A062319.
Array antidiagonal sums (row sums of the triangle) are A343657.
Dominated by A343658.
A000312 = n^n.
A007318 counts k-sets of elements of {1..n}.
A009998(n,k) = n^k (as an array, offset 1).
A059481 counts k-multisets of elements of {1..n}.
Cf. A000169, A000272, A002064, A002109, A066959, A143773, A146291, A176029, A251683, A282935, A326358, A327527, A334996, A343652.

Table[DivisorSigma[0,k^(n-k)],{n,10},{k,n}]

A(n, k) = numdiv(n^k); \\ Seiichi Manyama, May 15 2021

A(n,k) = A000005(A009998(n,k)), where A009998(n,k) = n^k is the interpretation as an array.

A(n,k) = Sum_{d|n} k^omega(d). - Seiichi Manyama, May 15 2021

A048472 Array T by antidiagonals, T(k,n)=(k+1)n2^(n-1)+1, n >= 0, k >= 1.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 13, 9, 4, 1, 33, 25, 13, 5, 1, 81, 65, 37, 17, 6, 1, 193, 161, 97, 49, 21, 7, 1, 449, 385, 241, 129, 61, 25, 8, 1, 1025, 897, 577, 321, 161, 73, 29, 9, 1, 2305, 2049, 1345, 769, 401, 193, 85, 33, 10, 1, 5121, 4609, 3073

Offset: 0

Clark Kimberling

n-th difference of (T(k,n),T(k,n-1),...,T(k,0)) is (k+1)n, for n=1,2,3,...; k=0,1,2,...

			Antidiagonals: {1}; {2,1}; {5,3,1}; ...

See A049069 for transposed array.
Row 1 = (1, 2, 5, 13, 33, ...) = A005183.
Row 2 = (1, 3, 9, 25, 65, ...) = A002064.

PARI
```
T(n,k)=if(n<0 || k<1,0,k*n*2^(n-1)+1)
```

Better description from Michael Somos

A100314 Number of 2 X n 0-1 matrices avoiding simultaneously the right angled numbered polyomino patterns (ranpp) (00;1), (01;0), (10;0) and (01;1).

Original entry on oeis.org

1, 4, 8, 14, 24, 42, 76, 142, 272, 530, 1044, 2070, 4120, 8218, 16412, 32798, 65568, 131106, 262180, 524326, 1048616, 2097194, 4194348, 8388654, 16777264, 33554482, 67108916, 134217782, 268435512, 536870970, 1073741884, 2147483710, 4294967360, 8589934658

Offset: 0

Sergey Kitaev, Nov 13 2004

An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2, j1 < j2 and these elements are in the same relative order as those in the triple (x,y,z). In general, the number of m X n 0-1 matrices in question is given by 2^m + 2^n + 2*(n*m-n-m).

Arthur H. Stroud, Approximate calculation of multiple integrals, Prentice-Hall, 1971.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Ronald Cools, Encyclopaedia of Cubature Formulas
S. Kitaev, On multi-avoidance of right angled numbered polyomino patterns, Integers: Electronic Journal of Combinatorial Number Theory 4 (2004), A21, 20pp.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. this sequence (m=2), A100315 (m=3), A100316 (m=4).
Cf. A000051, A000079, A001787, A002064, A005126, A005843.
Cf. A052548, A058331, A099003, A132750, A176691.
Row sums of A131830.

List([0..40],n->2^n+2*n); # Muniru A Asiru, Dec 21 2018

[2^n+2*n: n in [1..40]]; // Vincenzo Librandi, Oct 22 2011

a:= proc(n) 2^n + 2*n: end: seq(a(n),n=0..50); # Gary W. Adamson, Jul 20 2007

LinearRecurrence[{4,-5,2}, {1,4,8}, 34] (* Jean-François Alcover, Mar 19 2020 *)

makelist(2^n + 2*n, n, 0, 50); /* Franck Maminirina Ramaharo, Dec 19 2018 */

[2^n +2*n for n in range(41)] # G. C. Greubel, Feb 01 2023

a(n) = 2^n + 2*n.
From Gary W. Adamson, Jul 20 2007: (Start)
Binomial transform of (1, 3, 1, 1, 1, ...).
For n > 0, a(n) = 2*A005126(n-1). (End)
From R. J. Mathar, Jun 13 2008: (Start)
G.f.: 1 + 2*x*(2 -4*x +x^2)/((1-x)^2*(1-2*x)).
a(n+1)-a(n) = A052548(n). (End)
From Colin Barker, Oct 16 2013: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
G.f.: (1 - 3*x^2)/((1-x)^2*(1-2*x)). (End)
E.g.f.: exp(2*x) + 2*x*exp(x). - Franck Maminirina Ramaharo, Dec 19 2018
a(n) = A000079(n) + A005843(n). - Muniru A Asiru, Dec 21 2018

a(0)=1 prepended by Alois P. Heinz, Dec 21 2018

A048495 a(n) = (n-1)*2^n + 2.

Original entry on oeis.org

1, 2, 6, 18, 50, 130, 322, 770, 1794, 4098, 9218, 20482, 45058, 98306, 212994, 458754, 983042, 2097154, 4456450, 9437186, 19922946, 41943042, 88080386, 184549378, 385875970, 805306370, 1677721602, 3489660930, 7247757314

Offset: 0

Clark Kimberling

Binomial transform of 1 followed by the odd numbers (2n-1+2*0^n, or abs(A060747)). Binomial transform is A084643. - Paul Barry, Jun 09 2003
Total number of bits of all binary numbers less than 2^n (see example).
Total number of zero bits of all binary numbers less than 2^(n+1). - Olivier Gérard, Feb 25 2014.
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>2, 4>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element, and the fourth element is larger than the third element. - Sergey Kitaev, Dec 08 2020

			a(1)=2 : 0 1
a(2)=6 : 0 1 10 11
a(3)=18 : 0 1 10 11 100 101 110 111
a(4)=50 : 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111
...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alice L. L. Gao, Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao, Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Index entries for linear recurrences with constant coefficients, signature (5,-8,4).

a(n) = T(1, n), array T given by A048494.

Cf. A002064, A060747, A084643.

[(n-1)*2^n + 2: n in [0..30]]; // Vincenzo Librandi, Sep 25 2011

A048495:=n->(n-1)*2^n + 2; seq(A048495(n), n=0..30); # Wesley Ivan Hurt, Jun 29 2014

f[n_]:=(n-1)2^n+2;Array[f,29,0] (* Robert G. Wilson v, Jun 29 2014 *)
LinearRecurrence[{5,-8,4},{1,2,6},30] (* Harvey P. Dale, Jan 23 2015 *)

a(n) = (n-1)*2^n + 2; \\ Joerg Arndt, Feb 25 2014

Vec(-(4*x^2-3*x+1)/((x-1)*(2*x-1)^2) + O(x^100)) \\ Colin Barker, Jun 29 2014

a(n) - 1 = Sum_{i=0..n-1} (n-i) * 2^(n-i-1) = n*2^(n-1) + (n-1)*2^(n-2) + (n-2)*2^(n-3) + ... + 1*(2^0). - Matthew Erbst (matt(AT)erbst.org), Apr 19 2006
a(n) = 2 * A002064(n-1), n >= 1. - Omar E. Pol, Sep 30 2012
a(n) = a(n-1) + (2^n - 2^(n-1)) * n = a(n-1) + n*2^(n-1). - Olivier Gérard, Feb 25 2014
G.f.: -(4*x^2-3*x+1) / ((x-1)*(2*x-1)^2). - Colin Barker, Jun 29 2014
E.g.f.: exp(x)*(2 + exp(x)*(2*x - 1)). - Stefano Spezia, Feb 14 2025

Better description from John W. Layman, May 04 1999

A050920 Cullen primes: primes of the form n*2^n+1.

Original entry on oeis.org

3, 393050634124102232869567034555427371542904833

Offset: 1

N. J. A. Sloane, Dec 30 1999

The next term is too large to display here, having 1423 digits. See A005849.

			1 * 2^1 + 1 = 3, which is prime.
141 * 2^141 + 1 = 393050634124102232869567034555427371542904833, which is also prime.
The third Cullen prime is approximately 2.677114856136697933736444 * 10^1422.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B20.

Ray Ballinger, Cullen Primes: Definition and Status.
Chris K. Caldwell, Cullen Primes.
Eric Weisstein's World of Mathematics, Cullen Number.

See A005849 for the corresponding n.

Cf. A002064.

Select[Table[n * 2^n + 1, {n, 5000}], PrimeQ] (* Harvey P. Dale, Dec 14 2014 *)

a(n) = A002064(A005849(n)).

A064748 a(n) = n*10^n + 1.

Original entry on oeis.org

1, 11, 201, 3001, 40001, 500001, 6000001, 70000001, 800000001, 9000000001, 100000000001, 1100000000001, 12000000000001, 130000000000001, 1400000000000001, 15000000000000001, 160000000000000001, 1700000000000000001, 18000000000000000001, 190000000000000000001

Offset: 0

N. J. A. Sloane, Oct 19 2001

Number of digits in (10^n)^(10^n) in base 10. - Altug Alkan, Apr 25 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Paul Leyland, Cullen and Woodall numbers and their generalization to other bases.
Index entries for linear recurrences with constant coefficients, signature (21,-120,100).

Cf. A002064, A126431.

[ n*10^n+1: n in [0..20]]; // Vincenzo Librandi, Sep 16 2011

Table[n 10^n + 1, {n, 0, 18}] (* Michael De Vlieger, Apr 25 2016 *)

a(n) = n*10^n + 1; \\ Altug Alkan, Apr 25 2016

From Ilya Gutkovskiy, Apr 25 2016: (Start)
O.g.f.: (1 - 10*x + 90*x^2)/((1 - x)*(1 - 10*x)^2).
E.g.f.: (1 + 10*x*exp(9*x))*exp(x). (End)
From Elmo R. Oliveira, May 03 2025: (Start)
a(n) = 21*a(n-1) - 120*a(n-2) + 100*a(n-3).
a(n) = A126431(n) + 1. (End)

A343657 Sum of number of divisors of x^y for each x >= 1, y >= 0, x + y = n.

Original entry on oeis.org

1, 2, 4, 7, 12, 18, 27, 39, 56, 77, 103, 134, 174, 223, 283, 356, 445, 547, 666, 802, 959, 1139, 1344, 1574, 1835, 2128, 2454, 2815, 3213, 3648, 4126, 4653, 5239, 5888, 6608, 7407, 8298, 9288, 10385, 11597, 12936, 14408, 16025, 17799, 19746, 21882, 24221

Offset: 1

Gus Wiseman, Apr 29 2021

nonn

			The a(7) = 27 divisors:
  1  32  81  64  25  6  1
     16  27  32  5   3
     8   9   16  1   2
     4   3   8       1
     2   1   4
     1       2
             1

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Antidiagonal row sums (row sums of the triangle) of A343656.
Dominated by A343661.
A000005(n) counts divisors of n.
A000312(n) = n^n.
A007318(n,k) counts k-sets of elements of {1..n}.
A009998(n,k) = n^k (as an array, offset 1).
A059481(n,k) counts k-multisets of elements of {1..n}.
A343658(n,k) counts k-multisets of divisors of n.
Cf. A000169, A000272, A002064, A002109, A048691, A062319, A066959, A143773, A146291, A176029, A251683, A282935, A326358, A327527, A334996.

Total/@Table[DivisorSigma[0,k^(n-k)],{n,30},{k,n}]

a(n) = Sum_{k=1..n} A000005(k^(n-k)).

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cp's OEIS Frontend

A070940 Number of digits that must be counted from left to right to reach the last 1 in the binary representation of n.

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A003071 Sorting numbers: maximal number of comparisons for sorting n elements by list merging.

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A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A343656 Array read by antidiagonals where A(n,k) is the number of divisors of n^k.

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A048472 Array T by antidiagonals, T(k,n)=(k+1)*n*2^(n-1)+1, n >= 0, k >= 1.

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A100314 Number of 2 X n 0-1 matrices avoiding simultaneously the right angled numbered polyomino patterns (ranpp) (00;1), (01;0), (10;0) and (01;1).

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A048472 Array T by antidiagonals, T(k,n)=(k+1)n2^(n-1)+1, n >= 0, k >= 1.