A001792 -id:A001792 - cp's OEIS frontend

A336102 Number of inseparable multisets of size n covering an initial interval of positive integers.

0, 0, 1, 1, 3, 3, 8, 8, 20, 20, 48, 48, 112, 112, 256, 256, 576, 576, 1280, 1280, 2816, 2816, 6144, 6144, 13312, 13312, 28672, 28672, 61440, 61440, 131072, 131072, 278528, 278528, 589824, 589824, 1245184, 1245184, 2621440, 2621440, 5505024, 5505024, 11534336

Offset: 0

Gus Wiseman, Jul 08 2020

nonn

A multiset is separable if it has a permutation that is an anti-run, meaning there are no adjacent equal parts.
Alternatively, a multiset is separable if its greatest multiplicity is greater than the sum of its remaining multiplicities plus one.
Also the number of compositions of n whose greatest part is greater than the sum of its remaining parts plus one. For example, the a(2) = 1 through a(7) = 8 compositions are:
  (2)  (3)  (4)    (5)    (6)      (7)
            (1,3)  (1,4)  (1,5)    (1,6)
            (3,1)  (4,1)  (2,4)    (2,5)
                          (4,2)    (5,2)
                          (5,1)    (6,1)
                          (1,1,4)  (1,1,5)
                          (1,4,1)  (1,5,1)
                          (4,1,1)  (5,1,1)

			The a(2) = 1 through a(7) = 8 multisets:
  {11}  {111}  {1111}  {11111}  {111111}  {1111111}
               {1112}  {11112}  {111112}  {1111112}
               {1222}  {12222}  {111122}  {1111122}
                                {111123}  {1111123}
                                {112222}  {1122222}
                                {122222}  {1222222}
                                {122223}  {1222223}
                                {123333}  {1233333}

Michael De Vlieger, Table of n, a(n) for n = 0..6625
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4).

The strong (weakly decreasing multiplicities) case is A025065.
The bisection is A049610.
The separable version is A336103.
Sequences covering an initial interval are A000670.
Anti-run compositions are A003242.
Anti-run patterns are A005649.
Separable partitions are A325534.
Inseparable partitions are A325535.
Inseparable factorizations are A333487.
Anti-run compositions are ranked by A333489.
Heinz numbers of inseparable partitions are A335448.
Cf. A001792, A019472, A052841, A106351, A124767, A269134, A292884, A335433, A335126, A335452, A335548.

Table[Length[Join@@Permutations/@Select[IntegerPartitions[n],With[{mx=Max@@#},mx>1+Total[DeleteCases[#,mx,{1},1]]]&]],{n,0,15}]
(* Second program: *)
CoefficientList[Series[x^2*(1 - x) (x + 1)^2/(2 x^2 - 1)^2, {x, 0, 43}], x] (* Michael De Vlieger, Apr 07 2021 *)

a(2*n) = a(2*n + 1) = A049610(n + 1).
a(n) = 2^(n-1) - A336103(n).
A001792 repeated for n > 1. David A. Corneth, Jul 09 2020
From Chai Wah Wu, Apr 07 2021: (Start)
a(n) = 4*a(n-2) - 4*a(n-4) for n > 5.
G.f.: x^2*(1 - x)*(x + 1)^2/(2*x^2 - 1)^2. (End)

A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

Original entry on oeis.org

1, 6, 34, 178, 882, 4210, 19570, 89202, 400498, 1776754

Offset: 1

Gus Wiseman, Jun 23 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The set of binary indices of each nonnegative integer and its partition into anti-runs begins:
  0: {}      {{}}
  1: {1}     {{1}}
  2: {2}     {{2}}
  3: {1,2}   {{1},{2}}
  4: {3}     {{3}}
  5: {1,3}   {{1,3}}
  6: {2,3}   {{2},{3}}
  7: {1,2,3} {{1},{2},{3}}
The flattened version begins: {}, {1}, {2}, {1}, {2}, {3}, {1,3}, {2}, {3}, {1}, {2}, {3}. Of these sets, the first of length 2 is the sixth (starting with 0), so we have a(2) = 6.

For runs instead of anti-runs we have A001792.
The unflattened version is A052499.
Positions of first appearances in A384877, see A000120, A245562, A245563, A384890.
A023758 lists differences of powers of 2.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
q=Join@@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}];
Table[Position[q,i][[1,1]],{i,mnrm[q]}]

A033430 a(n) = 4*n^3.

Original entry on oeis.org

0, 4, 32, 108, 256, 500, 864, 1372, 2048, 2916, 4000, 5324, 6912, 8788, 10976, 13500, 16384, 19652, 23328, 27436, 32000, 37044, 42592, 48668, 55296, 62500, 70304, 78732, 87808, 97556, 108000, 119164, 131072, 143748, 157216, 171500, 186624, 202612, 219488

Offset: 0

Jeff Burch

2*a(n) = (2*n)^3 is a perfect cube.
Number of edges of the product of two complete bipartite graphs, each of order 2n, K_n,n x K_n,n - Roberto E. Martinez II, Jan 07 2002
This sequence is related to A049451 by a(n) = n*A049451(n) + sum( A049451(i), i=0..n-1 ) for n>0. - Bruno Berselli, Dec 19 2013
For n>=3, also the detour index of the n-gear graph. - Eric W. Weisstein, Dec 20 2017
For n > 0, this sequence can be obtained by summing consecutive blocks of odd numbers where the n-th block contains the next 2n odd numbers. - Marco Zárate, Jun 15 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..750
Frank Ellermann, Illustration of binomial transforms
Eric Weisstein's World of Mathematics, Detour Index
Eric Weisstein's World of Mathematics, Gear Graph
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A049451.
Cf. A000292, A007588.
Cf. A001844.

[4*n^3: n in [0..40]]; // Vincenzo Librandi, Apr 27 2011

4 Range[0, 40]^3 (* Harvey P. Dale, Sep 07 2016 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 4, 32, 108}, 40] (* Harvey P. Dale, Sep 07 2016 *)
Table[4 n^3, {n, 0, 20}] (* Eric W. Weisstein, Dec 20 2017 *)
CoefficientList[Series[(4 x (1 + 4 x + x^2))/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 20 2017 *)

a(n)=4*n^3 \\ Charles R Greathouse IV, Oct 07 2015

G.f. 4*x*(1+4*x+x^2)/ (x-1)^4. - R. J. Mathar, Feb 01 2011
From Ilya Gutkovskiy, May 25 2016: (Start)
E.g.f.: 4*x*(1 + 3*x + x^2)*exp(x).
Sum_{n>=1} 1/a(n) = zeta(3)/4. (End)
Product_{n>=1} a(n)/A280089(n) = Pi. - Daniel Suteu, Dec 26 2016
From Bruce J. Nicholson, Dec 07 2019: (Start)
a(n) = 24*A000292(n-1) + 4*n.
a(n) = 2*A007588(n) + 2*n. (End)
a(n) = Sum_{k=0..2*n-1} (2*n*(n-1)-2*k+1). - Sean A. Irvine, Jun 19 2025

A034007 First differences of A045891.

Original entry on oeis.org

1, 0, 2, 4, 9, 20, 44, 96, 208, 448, 960, 2048, 4352, 9216, 19456, 40960, 86016, 180224, 376832, 786432, 1638400, 3407872, 7077888, 14680064, 30408704, 62914560, 130023424, 268435456, 553648128, 1140850688, 2348810240

Offset: 0

Wolfdieter Lang

Let M_n be the n X n matrix m_(i,j) = 4 + abs(i-j) then det(M_n) = (-1)^(n+1)*a(n+2). - Benoit Cloitre, May 28 2002
Number of ordered pairs of (possibly empty) ordered partitions, each not beginning with 1. - Christian G. Bower, Jan 23 2004
If X_1, X_2, ..., X_n are 2-blocks of a (2n+4)-set X then, for n>=1, a(n+3) is the number of (n+1)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007
Except for an initial 1, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = (1 - S^2)^2; see A291000. - Clark Kimberling, Aug 24 2017
Conjecture 1: For compositions of n+k-1, a(n) is the number of runs of 1 of length k. Example: Among the compositions of 4+2-1 = 5, there are a(4) = 4 runs of two 1's: 3,[1,1]; [1,1],3; 1,2,[1,1] and [1,1],2,1. - Gregory L. Simay, Feb 18 2018
Conjecture 2: More generally, let R(n,m,k) = the number of runs of k m's in all compositions of n. Then R(n,m,k) = A045623(n-m*k) - 2*A045623(n-m*(k+1)) + A045623(n-m*(k+2)). For example, R(7,1,1) = A045623(6) - 2*A045623(5) + A045623(4) = 144 - 2*64 + 28 = 44 = a(7). - Gregory L. Simay, Feb 20 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Frank Ellermann, Illustration of binomial transforms.
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A045623, A045891, A291000.
Convolution of A034008 with itself.
Columns of A091613 converge to this sequence.

[1,0,2] cat [(n+5)*2^(n-4): n in [3..30]]; // G. C. Greubel, Sep 27 2022

Join[{1,0,2,a=4},Table[a=(2*(n+7)*a)/(n+6),{n,2,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2011 *)

a(n)=if(n<3,[1,0,2][n+1],(n+5)*2^(n-4)) \\ Charles R Greathouse IV, Jun 01 2011

[1,0,2]+[(n+5)*2^(n-4) for n in range(3,30)] # G. C. Greubel, Sep 27 2022

a(n) = Sum_{k = 0..n-3} (k+4)*binomial(n-3,k) for n >= 3. - N. J. A. Sloane, Jan 30 2008
a(n) = (n+5)*2^(n-4), n >= 3; a(0)=1, a(1)=0, a(2)=2.
G.f.: ((1-x)^2/(1-2*x))^2.
a(n) = Sum_{k=0..n} (k+1)*C(n-3,n-k). - Peter Luschny, Apr 20 2015
From Amiram Eldar, Jan 13 2021: (Start)
Sum_{n>=2} 1/a(n) = 512*log(2) - 74327/210.
Sum_{n>=2} (-1)^n/a(n) = 14579/70 - 512*log(3/2). (End)
E.g.f.: (1/16)*(11 - 12*x + 2*x^2 + (5+2*x)*exp(2*x)). - G. C. Greubel, Sep 27 2022

A209281 Start with first run [0,1] then, for n >= 2, the n-th run has length 2^n and is the concatenation of [a(1),a(2),...,a(2^n/2)] and [n-a(1),n-a(2),...,n-a(2^n/2)].

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 1, 2, 4, 3, 2, 3, 1, 2, 3, 2, 5, 4, 3, 4, 2, 3, 4, 3, 1, 2, 3, 2, 4, 3, 2, 3, 6, 5, 4, 5, 3, 4, 5, 4, 2, 3, 4, 3, 5, 4, 3, 4, 1, 2, 3, 2, 4, 3, 2, 3, 5, 4, 3, 4, 2, 3, 4, 3, 7, 6, 5, 6, 4, 5, 6, 5, 3, 4, 5, 4, 6, 5, 4, 5, 2, 3, 4, 3, 5, 4, 3

Offset: 1

Benoit Cloitre, Jan 16 2013

Also the sum of the odd bisection (odd-indexed parts) of the (n-1)-th composition in standard order, where the k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. Note that this sequence counts {} as composition number 1 (instead of the usual 0). For example, composition number 741 in standard order is (2,1,1,3,2,1), with odd bisection (2,1,2), so a(742) = 2 + 1 + 2 = 5. - Gus Wiseman, Aug 24 2021

			[0,1] -> [0,1] U [2-0,2-1] =
[0,1,2,1] -> [0,1,2,1] U [3-0,3-1,3-2,3-1] =
[0,1,2,1,3,2,1,2] etc.
From _Gus Wiseman_, Aug 08 2021: (Start)
As a triangle without the initial 0, row-lengths A000079:
  1
  2 1
  3 2 1 2
  4 3 2 3 1 2 3 2
  5 4 3 4 2 3 4 3 1 2 3 2 4 3 2 3
  6 5 4 5 3 4 5 4 2 3 4 3 5 4 3 4 1 2 3 2 4 3 2 3 5 4 3 4 2 3 4 3
(End)

Cf. A010060 (Thue-Morse), A103204 (indices of 1's).
Cf. A029837 (binary order), A000120 (binary weight), A006068 (inverse Gray), A272020 (bit positions).
Cf. A089215.
As a triangle: A000079 (row lengths), A001792 (row sums).
Other composition part sums: A124754. A346633.
Also the sum of row A346702(n-1) of A066099.
Cf. A346697 (on prime indices).

Table[Total[First/@Partition[Append[Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse,0],2]],{n,0,100}] (* Gus Wiseman, Aug 08 2021 *)

/* compute 2^15 terms */ v=[0,1];for(n=2,15,v=concat(v,vector(2^n/2,i,n-v[i]));a(n)=v[n];)

a(n) = n--; my(s=1,ns); while((ns=n>>s), n=bitxor(n,ns); s<<=1); hammingweight(n); \\ Kevin Ryde, May 14 2022

Let T(n)=A010060(n) then for n>=1 a(2n)=a(n)+1-T(n-1) and a(2n+1)=a(n+1)+T(n).
For n>=2 a(n) = a(ceiling(n/2))+T(n-1) hence:
a(n) = Sum_{k=0..ceiling(log(n-1)/log(2))} T(floor((n-1)/2^k)).
For k>=0 a(3*2^k+1)=1 (more precisely a(n)=1 iff n is in A103204), a(2^k+1)=k+1, a(5*2^k+1)=2, a(7*2^k+1)=k+2 etc.
From Gus Wiseman, Aug 18 2021: (Start)
a(n + 1) = (A029837(n) + A124754(n))/2.
a(n + 1) = A029837(n) - A346633(n).
a(n + 1) = A346633(n) - A124754(n).
a(n + 1) = A029837(A346702(n)).
(End)
From Kevin Ryde, May 14 2022: (Start)
a(n) = A000120(A006068(n-1)), binary weight of inverse binary Gray code.
a(n) = Sum_{k=1..A000120(n-1)} (-1)^(k-1) * A272020(n-1,k), alternating sum of 1-bit positions.
a(n) = A089215(n) - 1.
(End)

A055897 a(n) = n*(n-1)^(n-1).

Original entry on oeis.org

1, 2, 12, 108, 1280, 18750, 326592, 6588344, 150994944, 3874204890, 110000000000, 3423740047332, 115909305827328, 4240251492291542, 166680102383370240, 7006302246093750000, 313594649253062377472, 14890324713954061755186, 747581753430634213933056

Offset: 1

Christian G. Bower, Jun 12 2000

nonn

Total number of leaves in all labeled rooted trees with n nodes.
Number of endofunctions of [n] such that no element of [n-1] is fixed. E.g., a(3)=12: 123 -> 331, 332, 333, 311, 312, 313, 231, 232, 233, 211, 212, 213.
Number of functions f: {1, 2, ..., n} --> {1, 2, ..., n} such that f(1) != f(2), f(2) != f(3), ..., f(n-1) != f(n). - Warut Roonguthai, May 06 2006
Determinant of the n X n matrix ((2n, n^2, 0, ..., 0), (1, 2n, n^2, 0, ..., 0), (0, 1, 2n, n^2, 0, ..., 0), ..., (0, ..., 0, 1, 2n)). - Michel Lagneau, May 04 2010
For n > 1: a(n) = A240993(n-1) / A240993(n-2). - Reinhard Zumkeller, Aug 31 2014
Total number of points m such that f^(-1)(m) = {m}, (i.e., the preimage of m is the singleton set {m}) summed over all functions f:[n]->[n]. - Geoffrey Critzer, Jan 20 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Frank Ellermann, Illustration of binomial transforms
Index entries for sequences related to rooted trees

Cf. A003227, A003228, A055314, A055540, A055541, A060226, A118537.

Cf. A240993. Diagonal of A265583.

List([1..20], n-> n*(n-1)^(n-1)); # G. C. Greubel, Aug 10 2019

a055897 n = n * (n - 1) ^ (n - 1) -- Reinhard Zumkeller, Aug 31 2014

[n*(n-1)^(n-1): n in [1..20]] // Wesley Ivan Hurt, Jun 26 2014

A055897:=n->`if`(n=1,1,n*(n-1)^(n-1)); seq(A055897(n), n=1..20); # Wesley Ivan Hurt, Jun 26 2014

Join[{1},Table[n(n-1)^(n-1), {n,2,20}]] (* Harvey P. Dale, Jul 18 2011 *)

{a(n)=polcoeff(1/(1-n*x+x*O(x^n))^2, n)} \\ Paul D. Hanna, Dec 27 2012

[n*(n-1)^(n-1) for n in (1..20)] # G. C. Greubel, Aug 10 2019

E.g.f.: x/(1-T), where T=T(x) is Euler's tree function (see A000169).
a(n) = Sum_{k=1..n} A055302(n, k)*k.
a(n) = the n-th term of the (n-1)-th binomial transform of {1, 1, 4, 18, 96, ..., (n-1)*(n-1)!, ...} (cf. A001563). - Paul D. Hanna, Nov 17 2003
a(n) = (n-1)^(n-1) + Sum_{i=2..n} (n-1)^(n-i)*binomial(n-1, i-1)*(i-1) *(i-1)!. - Paul D. Hanna, Nov 17 2003
a(n) = [x^(n-1)] 1/(1 - (n-1)*x)^2. - Paul D. Hanna, Dec 27 2012
a(n) ~ exp(-1) * n^n. - Vaclav Kotesovec, Nov 14 2014

Additional comments from Vladeta Jovovic, Mar 31 2001 and Len Smiley, Dec 11 2001

A053464 a(n) = n*5^(n-1).

Original entry on oeis.org

0, 1, 10, 75, 500, 3125, 18750, 109375, 625000, 3515625, 19531250, 107421875, 585937500, 3173828125, 17089843750, 91552734375, 488281250000, 2593994140625, 13732910156250, 72479248046875, 381469726562500

Offset: 0

Barry E. Williams, Jan 13 2000

Arithmetic derivative of 5^n: a(n) = A003415(5^n). - Darrell Minor, Jul 21 2025

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Frank Ellermann, Illustration of binomial transforms
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 756.
Index entries for linear recurrences with constant coefficients, signature (10,-25).

Cf. A002697, A027471, A001787.

[n*(5^(n-1)): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

Join[{a=0,b=1},Table[c=10*b-25*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
Table[n*5^(n-1),{n,0,20}] (* or *) LinearRecurrence[{10,-25},{0,1},30] (* Harvey P. Dale, Jul 22 2014 *)

{a(n) = n*5^(n-1)}; /* Michael Somos, Sep 12 2005 */

[lucas_number1(n,10,25) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009

a(n) = Sum_{k=0..n} 5^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - Paul Barry, Oct 15 2004
a(n) = 10*a(n-1) - 25*a(n-2); n>1; a(0)=0, a(1)=1.
Fourth binomial transform of n (starting 0, 1, 10...) Convolution of powers of 5.
G.f.: x/(1-5*x)^2; E.g.f.: x*exp(5*x). - Paul Barry, Jul 22 2003
a(n) = - 25^n * a(-n) for all n in Z. - Michael Somos, Jun 26 2017
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 5*log(5/4).
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(6/5). (End)

More terms from James Sellers, Feb 02 2000

A062111 Upper-right triangle resulting from binomial transform calculation for nonnegative integers.

Original entry on oeis.org

0, 1, 1, 4, 3, 2, 12, 8, 5, 3, 32, 20, 12, 7, 4, 80, 48, 28, 16, 9, 5, 192, 112, 64, 36, 20, 11, 6, 448, 256, 144, 80, 44, 24, 13, 7, 1024, 576, 320, 176, 96, 52, 28, 15, 8, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 9, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 10

Offset: 0

Henry Bottomley, May 30 2001

From Philippe Deléham, Apr 15 2007: (Start)
This triangle can be found in the Laisant reference in the following form:
  .......................5...11..
  ...................4...9...20..
  ...............3...7..16...36..
  ...........2...5..12..28.......
  .......1...3...8..20..48.......
  ...0...1...4..12..32..80....... (End)
Triangle A152920 reversed. - Philippe Deléham, Apr 21 2009

			As a lower triangle (T(n, k)):
    0;
    1,   1;
    4,   3,   2;
   12,   8,   5,  3;
   32,  20,  12,  7,  4;
   80,  48,  28, 16,  9,  5;
  192, 112,  64, 36, 20, 11,  6;
  448, 256, 144, 80, 44, 24, 13, 7;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
F. Ellermann, Illustration of binomial transforms
C.-A. Laisant, Sur les tableaux de sommes - Nouvelles applications, Compt. Rendus de l'Association Francaise pour l'Avancement des Sciences, Aout 04 1893, pp. 206-216 (table given on p. 212).

Rows include (essentially) A001787, A001792, A034007, A045623, A045891.
Diagonals include (essentially) A001477, A005408, A008586, A008598, A017113.
Column sums are A058877.
Cf. A111297, A159694, A159695, A159696, A159697. - Philippe Deléham, Apr 21 2009
Cf. A058877, A073371, A130129, A152920.

[2^(n-k-1)*(n+k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 28 2022

Table[2^(n-k-1)*(n+k), {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 28 2022 *)

def A062111(n,k): return 2^(n-k-1)*(n+k)
flatten([[A062111(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 28 2022

A(n, k) = A(n, k-1) + A(n+1, k) if k > n with A(n, n) = n.
A(n, k) = (k+n)*2^(k-n-1) if k >= n.
T(2*n, n) = 3*n*2^(n-1) = 3*A001787(n). - Philippe Deléham, Apr 21 2009
From G. C. Greubel, Sep 28 2022: (Start)
T(n, k) = 2^(n-k-1)*(n+k) for 0 <= k <= n, n >= 0.
T(m*n, n) = 2^((m-1)*n-1)*(m+1)*A001477(n), m >= 1.
T(2*n-1, n-1) = A130129(n-1).
T(2*n+1, n-1) = 12*A001787(n).
Sum_{k=0..n} T(n, k) = A058877(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = 3*A073371(n-2), n >= 2.
T(n, k) = A152920(n, n-k). (End)

A053541 a(n) = n*10^(n-1).

Original entry on oeis.org

1, 20, 300, 4000, 50000, 600000, 7000000, 80000000, 900000000, 10000000000, 110000000000, 1200000000000, 13000000000000, 140000000000000, 1500000000000000, 16000000000000000, 170000000000000000

Offset: 1

Barry E. Williams, Jan 15 2000

This sequence gives the number of 1's (or any other nonzero digit) required to write all integers from 0 up to 10^n-1. - Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004 (improved by Bernard Schott, Nov 17 2022)

The corresponding number of 0's required to write all these integers from 0 up to 10^n-1 is A033714(n). - Bernard Schott, Nov 17 2022

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Frank Ellermann, Illustration of binomial transforms.
Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A001787, A033714, A038303, A053464, A053469, A081045, A094798.

List([1..20], n-> n*10^(n-1)) # G. C. Greubel, May 16 2019

[n*10^(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 06 2011

seq(n*10^(n-1), n = 1 .. 40); # Bernard Schott, Nov 17 2022

f[n_]:=n*10^(n-1);f[Range[40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011*)
LinearRecurrence[{20,-100},{1,20},20] (* Harvey P. Dale, Aug 08 2023 *)

a(n)=n*10^(n-1) \\ Charles R Greathouse IV, Dec 05 2011

[n*10^(n-1) for n in (1..20)] # G. C. Greubel, May 16 2019

a(n) = 20*a(n-1) - 100*a(n-2), with a(0)=0, a(1)=1, a(2)=20.
From Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004: (Start)
a(n) = 10*a(n-1) + 10*(n-1).
a(n) = Sum_{k=1..n} k*binomial(n,k)*9^(n-k).
a(n) = A094798(10^n - 1). (End)
From G. C. Greubel, May 16 2019: (Start)
G.f.: x/(1-10*x)^2.
E.g.f.: x*exp(10*x). (End)
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 10*log(10/9).
Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(11/10). (End)
a(n) = Sum_{k=1..n} A081045(k-1). - Bernard Schott, Nov 17 2022

Offset changed from 0 to 1 by Vincenzo Librandi, Jun 06 2011

A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

Original entry on oeis.org

0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432

Offset: 0

N. J. A. Sloane, James Sellers, Jan 20 2000

Define a(1)=0, a(2)=8 with 5*(a(1)^2) + 5*a(1) + 1 = j(1)^2 = 1^2 and 5*(a(2)^2) + 5*a(2) + 1 = j(2)^2 = 19^2. Then a(n) = a(n-2) + 8*sqrt(5*(a(n-1)^2) + 5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2) + 5*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey, Apr 03 2005
sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey, Apr 19 2005
a(n) is such that 5*a(n)^2 + 5*a(n) + 1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI, Apr 21 2005
From Russell Jay Hendel, Apr 25 2015: (Start)
We prove the two McGarvey-CAMI conjectures mentioned at the beginning of the Comments section. Let, as usual, F(n) = A000045(n), the Fibonacci numbers. In the sequel we indicate equations with upper case letters ((A), (B), (C), (D)) for ease of reference.
Then we must prove (A), 5*((F(6*n+3)-2)/4)^2 + 5*((F(6*n+3)-2)/4) + 1 = ((F(6*n+5)-F(6*n+1))/4)^2. Let m = 3*n+1 so that 6*n+1, 6*n+3, and 6*n+5 are 2*m-1, 2*m+1, and 2*m+3 respectively. Define G(m) = F(6*n+3) = F(2*m+1) = A001519(m+1), the bisected Fibonacci numbers. We can now simplify equation (A) by i) multiplying the LHS and RHS by 16, ii) expanding squares, and iii) gathering like terms. This shows proof of (A) equivalent to proving (B), 5*G(m)^2-4 = (G(m+1)-G(m-1))^2.
By Jarden's theorem (D. Jarden, Recurring sequences, 2nd ed. Jerusalem, Riveon Lematematika, (1966)), if {H(n)}{n >=1} is any recursive sequence satisfying (C), H(n)=3H(n-1)-H(n-2), then {H(n)}^2{n >=1} is also a recursive sequence satisfying (D), H(n)^2=8H(n-1)^2-8H(n-2)^2+H(n-3)^2. As noted in the Formula section of A001519, {G(m)}_{m >= 1} satisfies (C).
Proof of (B) is now straightforward. Since {G(m)}{m >=1} satisfies (C), it follows that {G(m)^2}{m >=1} satisfies (D), and therefore, {5G(m)^2-4}_{m >=1} also satisfies (D).
Similarly, since {G(m)}{m >=1} satisfies (C), it follows that both {G(m+1)}{m >=1}, {G(m-1)}{m >=1} and their difference {G(m+1)-G(m-1)}{m >=1} satisfy (C), and therefore {G(m+1)-G(m-1)}^2_{m >=1} satisfies (D).
But then the LHS and RHS of (B) are equal for m=1,2,3 and satisfy the same recursion, (D). Hence the LHS and RHS of (B) are equal for all m. This completes the proof. (End)

G. C. Greubel, Table of n, a(n) for n = 0..790
F. Ellermann, Illustration of binomial transforms
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A049629.

Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.

List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # G. C. Greubel, May 16 2019

[(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // Vincenzo Librandi, Apr 20 2011

A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
seq(A053606(n),n=0..30) ;

Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

a(n)=fibonacci(6*n+3)\4 \\ Charles R Greathouse IV, Jul 02 2013

[(fibonacci(6*n+3)-2)/4 for n in (0..30)] # G. C. Greubel, May 16 2019

a(n) = 8*A049664(n).
a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - Richard Choulet, Aug 30 2007
G.f.: 8*x/((1-x)*(1-18*x+x^2)). - Richard Choulet, Oct 09 2007
a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - Gary Detlefs, Dec 07 2010
a(n) = Sum_{k=0..n} A134492(k). - Gary Detlefs, Dec 07 2010
a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - Gary Detlefs, Dec 08 2010

cp's OEIS Frontend

A336102 Number of inseparable multisets of size n covering an initial interval of positive integers.

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A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

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A033430 a(n) = 4*n^3.

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A034007 First differences of A045891.

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A209281 Start with first run [0,1] then, for n >= 2, the n-th run has length 2^n and is the concatenation of [a(1),a(2),...,a(2^n/2)] and [n-a(1),n-a(2),...,n-a(2^n/2)].

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A055897 a(n) = n*(n-1)^(n-1).

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A053464 a(n) = n*5^(n-1).

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