A055272 -id:A055272 - cp's OEIS frontend

A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

1, -1, 1, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1

Offset: 0

Philippe Deléham, Nov 02 2009

Riordan array (1-x,1) read by rows; Riordan inverse is (1/(1-x),1). Columns have g.f. (1-x)x^k. Diagonal sums are A033999. Unsigned version in A097806.
Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = -1, T(n,k) = 0, k > 2. - Boris Putievskiy, Jan 17 2013
Finite difference operator (pair difference): left multiplication by T of a sequence arranged as a column vector gives a running forward difference, a(k+1)-a(k), or first finite difference (modulo sign), of the elements of the sequence. T^n gives the n-th finite difference (mod sign). T is the inverse of the summation matrix A000012 (regarded as lower triangular matrices). - Tom Copeland, Mar 26 2014

			Triangle begins:
   1;
  -1,  1;
   0, -1,  1;
   0,  0, -1,  1;
   0,  0,  0, -1,  1;
   0,  0,  0,  0, -1,  1; ...
Row number r (r>4) contains (r-2) times '0', then '-1' and '1'.
From _Boris Putievskiy_, Jan 17 2013: (Start)
The start of the sequence as a table:
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  ...
(End)

Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Cf. A000012, A007318, A029653, A097805, A118800, A130595, A156644.

A167374 := proc(n,k)
    if k> n or k < n-1 then
        0;
    elif k = n then
        1;
    else
        -1 ;
    end if;
end proc: # R. J. Mathar, Sep 07 2016

Table[PadLeft[{-1, 1}, n], {n, 13}] // Flatten (* or *)
MapIndexed[Take[#1, First@ #2] &, CoefficientList[Series[(1 - x)/(1 - x y), {x, 0, 12}], {x, y}]] // Flatten (* Michael De Vlieger, Nov 16 2016 *)
T[n_, k_] := If[ k<0 || k>n, 0, Boole[n==k] - Boole[n==k+1]]; (* Michael Somos, Oct 01 2022 *)

{T(n, k) = if( k<0 || k>n, 0, (n==k) - (n==k+1))}; /* Michael Somos, Oct 01 2022 */

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A055268(n), A055276(n) for x = 1,2,3,4,5,6,7,8,9,10,11 respectively .
From Boris Putievskiy, Jan 17 2013: (Start)
a(n) = floor((A002260(n)+2)/(A003056(n)+2))*(-1)^(A002260(n)+A003056(n)+1),  n>0.
a(n) = floor((i+2)/(t+2))*(-1)^(i+t+1), n > 0, where
i = n - t*(t+1)/2,
t = floor((-1 + sqrt(8*n-7))/2). (End)
T*A000012 = Identity matrix. T*A007318 = A097805. T*(A007318)^(-1)= signed A029653. - Tom Copeland, Mar 26 2014
G.f.: (1-x)/(1-x*y). - R. J. Mathar, Aug 11 2015
T = A130595*A156644 = M*T^(-1)*M = M*A000012*M, where M(n,k) = (-1)^n A130595(n,k). Note that M = M^(-1). Cf. A118800 and A097805. - Tom Copeland, Nov 15 2016

A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 16, 48, 56, 32, 9, 1, 32, 112, 160, 120, 50, 11, 1, 64, 256, 432, 400, 220, 72, 13, 1, 128, 576, 1120, 1232, 840, 364, 98, 15, 1, 256, 1280, 2816, 3584, 2912, 1568, 560, 128, 17, 1, 512, 2816, 6912, 9984, 9408, 6048, 2688, 816, 162, 19, 1

Offset: 0

Philippe Deléham, Nov 13 2011

Riordan array ((1-x)/(1-2x),x/(1-2x)).
Product A097805*A007318 as infinite lower triangular arrays.
Product A193723*A130595 as infinite lower triangular arrays.
T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - Geoffrey Critzer, Nov 18 2012
Apparently, rows of this array are unsigned diagonals of A028297. - Tom Copeland, Oct 11 2014
Unsigned A118800, so my conjecture above is true. - Tom Copeland, Nov 14 2016

			Triangle begins:
   1
   1,   1
   2,   3,   1
   4,   8,   5,   1
   8,  20,  18,   7,   1
  16,  48,  56,  32,   9,   1
  32, 112, 160, 120,  50,  11,   1

Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Cf. A007318, A028297, A059260, A097805, A118801, A130595.

nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n

T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.

G.f.: (1-x)/(1-(2+y)*x).

T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.

T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - Tom Copeland, Nov 15 2016

T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - Tom Copeland, Nov 17 2016

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 24 2018

A193723 Mirror of the fusion triangle A193722.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 18, 21, 8, 1, 54, 81, 45, 11, 1, 162, 297, 216, 78, 14, 1, 486, 1053, 945, 450, 120, 17, 1, 1458, 3645, 3888, 2295, 810, 171, 20, 1, 4374, 12393, 15309, 10773, 4725, 1323, 231, 23, 1, 13122, 41553, 58320, 47628, 24948, 8694, 2016, 300, 26, 1

Offset: 0

Clark Kimberling, Aug 04 2011

A193723 is obtained by reversing the rows of the triangle A193722.
Triangle T(n,k), read by rows, given by [2,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
From Philippe Deléham, Nov 14 2011: (Start)
Riordan array ((1-x)/(1-3x), x/(1-3x)).
Product A200139*A007318 as infinite lower triangular arrays. (End)

			First six rows:
    1;
    2,   1;
    6,   5,   1;
   18,  21,   8,   1;
   54,  81,  45,  11,   1;
  162, 297, 216,  78,  14,   1;

Cf. A084938, A193722, A052924 (antidiagonal sums), Diagonals: A000012, A016789, A081266, Columns: A025192, A081038.

z = 9; a = 1; b = 1; c = 1; d = 2;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)

Write w(n,k) for the triangle at A193722.  The triangle at A193723 is then given by w(n,n-k).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,1)=1 and T(1,0)=2. - Philippe Deléham, Oct 05 2011
From Philippe Deléham, Nov 14 2011: (Start)
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for x=-2,-1,0,1,2,3,4,5,6,7,8,9 respectively.
T(n,k) = Sum_{j>=0} T(n-1-j,k-1)*3^j.
G.f.: (1-x)/(1-(3+y)*x). (End)

A085388 First differences of n^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 6, 4, 0, 1, 4, 12, 18, 8, 0, 1, 5, 20, 48, 54, 16, 0, 1, 6, 30, 100, 192, 162, 32, 0, 1, 7, 42, 180, 500, 768, 486, 64, 0, 1, 8, 56, 294, 1080, 2500, 3072, 1458, 128, 0, 1, 9, 72, 448, 2058, 6480, 12500, 12288, 4374, 256, 0, 1, 10, 90, 648

Offset: 1

Paul Barry, Jun 30 2003

T(n,k) is the number of k-digit numbers in base n; n,k >= 2. - Mohammed Yaseen, Nov 11 2022

			Rows begin
  1,   0,   0,   0,   0, ...
  1,   1,   2,   4,   8, ...
  1,   2,   6,  18,  54, ...
  1,   3,  12,  48, 192, ...
  1,   4,  20, 100, 500, ...

Rows include A011782, A025192, A002001, A005054, A052934, A055272, A055274, A055275, A052268, A055276.
Diagonals include A053506, A085389, A085390.
Row-wise binomial transform is A083064.

T(n,k) = (n-1)*n^(k-1) + 0^k/n. - Corrected by Mohammed Yaseen, Nov 11 2022

T(n,0) = 1; T(n,k) = n^k - n^(k-1) for k >= 1. - Mohammed Yaseen, Nov 11 2022

Offset corrected by Mohammed Yaseen, Nov 11 2022

A055270 a(n) = 7a(n-1) + (-1)^n binomial(2,2-n) with a(-1)=0.

Original entry on oeis.org

1, 5, 36, 252, 1764, 12348, 86436, 605052, 4235364, 29647548, 207532836, 1452729852, 10169108964, 71183762748, 498286339236, 3488004374652, 24416030622564, 170912214357948, 1196385500505636, 8374698503539452, 58622889524776164, 410360226673433148, 2872521586714032036

Offset: 0

Barry E. Williams, May 10 2000

For n >= 2, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4,5,6,7} such that for fixed, different x_1, x_2 in {1,2,...,n} and fixed y_1, y_2 in {1,2,3,4,5,6,7} we have f(x_1) <> y_1 and f(x_2) <> y_2. - Milan Janjic, Apr 19 2007

a(n) is the number of generalized compositions of n when there are 6*i-1 different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (7).

Cf. A055272 (first differences of 7^n (A000420)).

[1,5] cat [36*7^(n-2): n in [2..30]]; // G. C. Greubel, Mar 16 2020

A055270:= n-> `if`(n<2, 4*n+1, 36*7^(n-2)); seq(A055270(n), n=0..30); # G. C. Greubel, Mar 16 2020

Join[{1,5},NestList[7#&,36,20]] (* Harvey P. Dale, Sep 04 2017 *)

[1,5]+[36*7^(n-2) for n in (2..30)] # G. C. Greubel, Mar 16 2020

a(n) = 6^2 * 7^(n-2), n >= 2 with a(0)=1, a(1)=5.
G.f.: (1-x)^2/(1-7*x).
a(n) = Sum_{k=0..n} A201780(n,k)*5^k. - Philippe Deléham, Dec 05 2011
E.g.f.: (13 - 7*x + 36*exp(7*x))/49. - G. C. Greubel, Mar 16 2020

Terms a(20) onward added by G. C. Greubel, Mar 16 2020

A329573 For all n >= 1, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 4, 9, 10, 27, 14, 20, 33, 34, 69, 39, 28, 40, 13, 19, 70, 31, 43, 180, 220, 61, 36, 66, 91, 127, 7, 12, 5, 102, 186, 11, 6, 25, 18, 55, 41, 42, 48, 65, 72, 59, 38, 125, 24, 29, 35, 54, 32, 47, 77, 164, 26, 407, 15, 116, 63, 75, 404, 416, 8, 215, 45, 56, 183, 23, 134, 206, 17, 44, 50

Offset: 1

M. F. Hasler, Feb 09 2020

nonn

That is, there are 12 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
This is the theoretical maximum: there can't be more than 12 primes in pairwise sums of 7 distinct numbers > 1. See the wiki page for more details.
Conjectured to be a permutation of the positive integers. See A329572 for the nonnegative variant (same definition but with n >= 0 and terms >= 0), leading to a quite different sequence.
For a(5) and a(6) one must forbid values up to 8 in order to be able to find a solution for a(7), but from then on, the greedy choice gives the correct solution, at least for several hundred terms. Small values appearing late are a(30) = 5, a(34) = 6, a(28) = 7, a(62) = 8.

			Up to and including the 6th term, there is no constraint other than not using a term more than once, since it is impossible to have more than 12 primes as pairwise sums of 6 numbers. So one would first try to use the lexicographically smallest possible choice a(1..6) =?= (1, 2, ..., 6). But then one would have only 7 pairs (i,j) such that a(i) + a(j) is prime, 1 <= i < j <= 6. So one would need 12 - 7 = 5 more primes in {1, 2, ..., 6} + a(7), which is impossible. One can check that even a(1..5) =?= (1,...,5) does not allow one to find a(6) and a(7) in order to have 12 prime sums a(i) + a(j), 1 <= i < j <= 7. Nor is it possible to find a solution with a(5) equal to 6 or 7 or 8. One finds that a(5) = 9, and a(6) = 10, are the smallest possible choices for which a(7) can be found as to satisfy the requirement. In that case, a(7) = 27 is the smallest possible solution, which yields the 12 prime sums 1+2, 2+3, 1+4, 3+4, 2+9, 4+9, 1+10, 3+10, 9+10, 2+27, 4+27, 10+27.
Now, to satisfy the definition of the sequence for n = 2, we drop the initial 1 from the set of consecutive terms, and search for a(8) producing the same number of additional primes together with {2, 3, 4, 9, 10, 27} as did a(1) = 1, namely 3. We see that a(8) = 14 is the smallest possibility. And so on.
It seems that once a(5) and a(6) are chosen, one may always take the smallest possible choice for the next term without ever again running into difficulty. This is in strong contrast to the (exceptional) case of the variant where we require 10 prime sums among 7 consecutive terms, cf. sequence A329574.

M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

Cf. A055272 (analog starting with a(0)=0), A055265 & A128280 (1 prime using 2 terms), A055266 & A253074 (0 primes using 2 terms), A329405 - A329416, A329425, A329333, A329449 - A329456, A329563 - A329581.

{A329573(n,show=0,o=1,N=12,M=6,D=[5,9,6,10],p=[],u=o,U)=for(n=o+1,n, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); D&& D[1]==n&& [o=D[2],D=D[3..-1]]&& next; my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); for(k=u,oo,bittest(U,k-u)|| min(c-#[0|p<-p,isprime(p+k)],#p>=M)|| [o=k,break]));show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. See the wiki page for more.

A196731 Expansion of g.f. (1-x)/(1-12*x).

Original entry on oeis.org

1, 11, 132, 1584, 19008, 228096, 2737152, 32845824, 394149888, 4729798656, 56757583872, 681091006464, 8173092077568, 98077104930816, 1176925259169792, 14123103110037504, 169477237320450048, 2033726847845400576, 24404722174144806912, 292856666089737682944, 3514279993076852195328

Offset: 0

Philippe Deléham, Oct 05 2011

Index entries for linear recurrences with constant coefficients, signature (12).

Cf. A002001, A005054, A025192, A052934, A055272, A055274, A055276, A193722.

a:= n-> ceil(11*12^(n-1)):
seq(a(n), n=0..20);  # Alois P. Heinz, Mar 25 2025

Join[{1},NestList[12#&,11,20]] (* Harvey P. Dale, Sep 19 2018 *)

a(n)=if(n,11*12^n--,1) \\ Charles R Greathouse IV, Oct 05 2011

a(n)=(11+!n)*12^(n-1)  \\ M. F. Hasler, Oct 05 2011

a(n) = Sum_{k=0..n} A193722(n,k)*9^(n-k).
a(n+1) = 12*a(n) for n > 0. - M. F. Hasler, Oct 05 2011
From Elmo R. Oliveira, Mar 18 2025: (Start)
a(n) = 11*12^(n-1) with a(0)=1.
E.g.f.: (11*exp(12*x) + 1)/12. (End)

More terms from Elmo R. Oliveira, Mar 25 2025

A334603 Period of the fraction 1/11^n for n >= 1.

Original entry on oeis.org

2, 22, 242, 2662, 29282, 322102, 3543122, 38974342, 428717762, 4715895382, 51874849202, 570623341222, 6276856753442, 69045424287862, 759499667166482, 8354496338831302, 91899459727144322, 1010894056998587542, 11119834626984462962, 122318180896829092582

Offset: 1

Bernard Schott, May 07 2020

Conjecture proposed by the authors in References page 205: if p is a prime with gcd(p,30) = 1 and if the period of 1/p is m then the period of 1/p^n is m*p^(n-1).

			1/121 = 0. 0082644628099173553719 0082644628099173553719 ... with periodic part {0082644628099173553719}, whose length is 22 digits, so a(2) = 22.

J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 346 pp. 50, 204-205, Ellipses, Paris 2004.

Cf. period of fractions: A051626 (1/n), A133494 (1/3^n), A055272 (1/7^n).

Cf. A001020 (11^n).

MultiplicativeOrder[10, 11^#] & /@ Range[20] (* Giovanni Resta, May 07 2020 *)

a(n) = znorder(Mod(10, 11^n)); \\ Michel Marcus, May 09 2020

a(n) = 2 * 11^(n-1) [conjectured, see comments].

a(n) = A051626(A001020(n)).

More terms from Giovanni Resta, May 07 2020

Showing 1-8 of 8 results.

cp's OEIS Frontend

A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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A193723 Mirror of the fusion triangle A193722.

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A085388 First differences of n^k.

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A055270 a(n) = 7*a(n-1) + (-1)^n * binomial(2,2-n) with a(-1)=0.

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A329573 For all n >= 1, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct positive numbers.

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A196731 Expansion of g.f. (1-x)/(1-12*x).

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A055270 a(n) = 7a(n-1) + (-1)^n binomial(2,2-n) with a(-1)=0.