A081294 -id:A081294 - cp's OEIS frontend

A053661 For n > 1: if n is present, 2n is not.

1, 2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 57, 59, 60, 61, 63, 65, 67, 68, 69, 71, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 89, 91, 92, 93, 95, 97, 99, 100, 101, 103, 105

Offset: 1

Jeevan Chana Rai (Karanjit.Rai(AT)btinternet.com), Feb 16 2000

The Name line gives a property of the sequence, not a definition. The sequence can be defined simultaneously with b(n) := A171945(n) via a(n) = mex{a(i), b(i) : 0 <= i < n} (n >= 0}, b(n)=2a(n). The two sequences are complementary, hence A053661 is identical to A171944 (except for the first terms). Furthmore, A053661 is the same as A003159 except for the replacement of vile by dopey powers of 2. - Aviezri S. Fraenkel, Apr 28 2011
For n >= 2, either n = 2^k where k is odd or n = 2^k*m where m > 1 is odd and k is even (found by Kirk Bresniker and Stan Wagon). [Robert Israel, Oct 10 2010]
Subsequence of A175880; A000040, A001749, A002001, A002042, A002063, A002089, A003947, A004171 and A081294 are subsequences.

Robert Israel, Table of n, a(n) for n = 1..10000
Aviezri S. Fraenkel, The vile, dopey, evil and odious game players, Discrete Math. 312 (2012), no. 1, 42-46.

Essentially identical to A171944 and the complement of A171945.

a053661 n = a053661_list !! (n-1)
a053661_list = filter (> 0) a175880_list -- Reinhard Zumkeller, Feb 09 2011

N:= 1000: # to get all terms <= N
sort([1,seq(2^(2*i+1),i=0..(ilog2(N)-1)/2), seq(seq(2^(2*i)*(2*j+1),j=1..(N/2^(2*i)-1)/2),i=0..ilog2(N)/2)]); # Robert Israel, Jul 24 2015

Clear[T]; nn = 105; T[n_, k_] := T[n, k] = If[n < 1 || k < 1, 0, If[n == 1 || k == 1, 1, If[k > n, T[k, n], If[n > k, T[k, Mod[n, k, 1]], -Product[T[n, i], {i, n - 1}]]]]]; DeleteCases[Table[If[T[n, n] == -1, n, ""], {n, 1, nn}], ""] (* Mats Granvik, Aug 25 2012 *)

More terms from James Sellers, Feb 22 2000

A099142 a(n) = 6^n * T(n, 4/3) where T is the Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 8, 92, 1184, 15632, 207488, 2757056, 36643328, 487039232, 6473467904, 86042074112, 1143628341248, 15200538791936, 202038000386048, 2685388609667072, 35692849740775424, 474411605904392192

Offset: 0

Paul Barry, Sep 30 2004

In general, r^n * T(n,(r+2)/r) has g.f. (1-(r+2)*x)/(1-2*(r+2)*x + r^2*x^2), e.g.f. exp((r+2)*x)*cosh(2*sqrt(r+1)*x), a(n) = Sum_{k=0..n} (r+1)^k*binomial(2n,2k) and a(n) = (1+sqrt(r+1))^(2*n)/2 + (1-sqrt(r+1))^(2*n)/2.

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (16,-36).

Column k=7 of A333988.

Cf. A001541, A081294, A083884, A090965, A099140, A099141.

LinearRecurrence[{16,-36},{1,8},20] (* Harvey P. Dale, Mar 09 2018 *)

a(n) = 6^n*polchebyshev(n, 1, 4/3); \\ Michel Marcus, Sep 08 2019

G.f.: (1-8*x)/(1-16*x+36*x^2);
E.g.f.: exp(8*x)*cosh(2*sqrt(7)*x).
a(n) = 6^n * T(n, 8/6) where T is the Chebyshev polynomial of the first kind.
a(n) = Sum_{k=0..n} 7^k * binomial(2n, 2k).
a(n) = (1+sqrt(7))^(2*n)/2 + (1-sqrt(7))^(2*n)/2.
a(0)=1, a(1)=8, a(n) = 16*a(n-1) - 36*a(n-2) for n > 1. - Philippe Deléham, Sep 08 2009

A304411 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)*k_j).

Original entry on oeis.org

1, 3, 4, 6, 6, 12, 8, 9, 8, 18, 12, 24, 14, 24, 24, 12, 18, 24, 20, 36, 32, 36, 24, 36, 12, 42, 12, 48, 30, 72, 32, 15, 48, 54, 48, 48, 38, 60, 56, 54, 42, 96, 44, 72, 48, 72, 48, 48, 16, 36, 72, 84, 54, 36, 72, 72, 80, 90, 60, 144, 62, 96, 64, 18, 84, 144, 68, 108, 96, 144, 72, 72

Offset: 1

Ilya Gutkovskiy, May 12 2018

			a(24) = a(2^3*3) = (2 + 1)*3 * (3 + 1)*1 = 36.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Ilya Gutkovskiy, Scatter plot of a(n) up to n=50000 .
Index entries for sequences computed from exponents in factorization of n.
Index entries for sequences computed from indices in prime factorization.

Cf. A000005, A000026, A000040, A000203, A003959, A001615, A003961, A005117, A005361, A007947, A008864, A045965, A048250, A064478, A081294 (numbers n such that a(n) is odd), A181797, A304407, A304408, A304409, A304412.

Cf. A072691, A330596.

a[n_] := Times @@ ((#[[1]] + 1) #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 72}]
Table[Total[Select[Divisors[n], SquareFreeQ]] DivisorSigma[0, n/Last[Select[Divisors[n], SquareFreeQ]]], {n, 72}]

a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); (p+1)*e)} \\ Andrew Howroyd, Jul 24 2018

a(n) = A005361(n)*A048250(n) = A000005(n/A007947(n))*A000203(A007947(n)).
a(p^k) = (p + 1)*k where p is a prime and k > 0.
a(n) = Product_{p|n} (p + 1) if n is a squarefree (A005117).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A072691 * A330596 = 0.6156455744... . - Amiram Eldar, Nov 30 2022

A333988 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of (1-(k+1)x) / (1-2(k+1)x+((k-1)x)^2).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 8, 1, 1, 4, 17, 32, 1, 1, 5, 28, 99, 128, 1, 1, 6, 41, 208, 577, 512, 1, 1, 7, 56, 365, 1552, 3363, 2048, 1, 1, 8, 73, 576, 3281, 11584, 19601, 8192, 1, 1, 9, 92, 847, 6016, 29525, 86464, 114243, 32768, 1, 1, 10, 113, 1184, 10033, 62976, 265721, 645376, 665857, 131072, 1

Offset: 0

Seiichi Manyama, Sep 04 2020

			Square array begins:
  1,   1,    1,     1,     1,     1, ...
  1,   2,    3,     4,     5,     6, ...
  1,   8,   17,    28,    41,    56, ...
  1,  32,   99,   208,   365,   576, ...
  1, 128,  577,  1552,  3281,  6016, ...
  1, 512, 3363, 11584, 29525, 62976, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Column k=0..9 give A000012, A081294, A001541, A090965, A083884, A099140, A099141, A099142, A165224, A026244.
Main diagonal gives A333990.
Cf. A009999, A307883, A337389, A333989.

T[n_, 0] := 1; T[n_, k_] := Sum[k^j * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Sep 04 2020 *)

{T(n, k) = sum(j=0, n, k^j*binomial(2*n, 2*j))}

T(n,k) = Sum_{j=0..n} k^j * binomial(2*n,2*j).

T(0,k) = 1, T(1,k) = k+1 and T(n,k) = 2 * (k+1) * T(n-1,k) - (k-1)^2 * T(n-2,k) for n>1.

A115716 A divide-and-conquer sequence.

Original entry on oeis.org

1, 1, 3, 1, 3, 1, 11, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 683, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1

Offset: 0

Paul Barry, Jan 29 2006

			G.f. = 1 + x + 3*x^2 + x^3 + 3*x^4 + x^5 + 11*x^6 + x^7 + 3*x^8 + x^9 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..16381
R. Stephan, Divide-and-conquer generating functions. I. Elementary sequences , arXiv:math/0307027 [math.CO], 2003.

Partial sums are A032925.
Row sums of number triangle A115717.
Bisection: A276390.
Cf. A007583, A081294, A091090.
See A276391 for a closely related sequence.

a:= proc(n) option remember; `if`(n=0, 1,
      `if`(n::odd, 1, 4*a(n/2-1)-1))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Sep 07 2016

a[n_] := a[n] = If[n == 0, 1, If[OddQ[n], 1, 4*a[n/2-1]-1]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 25 2017, after Alois P. Heinz *)

{a(n) = if( n<1, n==0, n%2, 1, 4 * a(n/2-1) - 1)}; /* Michael Somos, Sep 07 2016 */

The g.f. G(x) satisfies G(x)-4*x^2*G(x^2)=(1+2*x)/(1+x). - Argument and offset corrected by Bill Gosper, Sep 07 2016
G.f.: 1/(1-x) + Sum_{k>=0} ((4^k-0^k)/2) *x^(2^(k+1)-2) /(1-x^(2^k)). - corrected by R. J. Mathar, Sep 07 2016
a(n)=A007583(A091090(n+1)-1). - Adapted to new offset by R. J. Mathar, Sep 07 2016
a(0) = 1, a(2*n + 1) = 1 for n>=0. a(2*n + 2) = 4*a(n) - 1 for n>=0. - Michael Somos, Sep 07 2016

A134683 Expansion of 1+x(2+3x)/(1-4*x^2).

Original entry on oeis.org

1, 2, 3, 8, 12, 32, 48, 128, 192, 512, 768, 2048, 3072, 8192, 12288, 32768, 49152, 131072, 196608, 524288, 786432, 2097152, 3145728, 8388608, 12582912, 33554432, 50331648, 134217728, 201326592, 536870912, 805306368, 2147483648, 3221225472

Offset: 0

Paul Curtz, Jan 26 2008, Feb 09 2008

A002001 interleaved with A081294. Gary W. Adamson, Jul 08 2012

Index entries for linear recurrences with constant coefficients, signature (0,4).

Cf. A001045, A045623, A131577, A002001, A081294, A000302.

A134683 := proc(n)
    if n =0 then
        1 ;
    else
        -(-1)^n*A131577(n-1)+2*procname(n-1) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012

CoefficientList[Series[1 + x*(2 + 3*x)/(1 - 4*x^2), {x, 0, 32}], x] (* Amiram Eldar, Aug 18 2022 *)

a(n) = 2*a(n-1)-(-1)^n*A131577(n-1), n>0.
a(n) = 4*a(n-2), n>2. Gary W. Adamson, Jul 08 2012
a(n) = 2^(n-3)*(7-(-1)^n), n>0. - R. J. Mathar, Jul 22 2012
Sum_{n>=0} 1/a(n) = 19/9. - Amiram Eldar, Aug 18 2022

Edited by N. J. A. Sloane, Feb 20 2008

A304407 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)*k_j).

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 3, 4, 4, 10, 4, 12, 6, 8, 4, 16, 4, 18, 8, 12, 10, 22, 6, 8, 12, 6, 12, 28, 8, 30, 5, 20, 16, 24, 8, 36, 18, 24, 12, 40, 12, 42, 20, 16, 22, 46, 8, 12, 8, 32, 24, 52, 6, 40, 18, 36, 28, 58, 16, 60, 30, 24, 6, 48, 20, 66, 32, 44, 24, 70, 12, 72, 36, 16

Offset: 1

Ilya Gutkovskiy, May 12 2018

			a(60) = a(2^2*3*5) = (2 - 1)*2 * (3 - 1)*1 * (5 - 1)*1 = 16.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Ilya Gutkovskiy, Scatter plot of a(n) up to n=50000.
Index entries for sequences computed from exponents in factorization of n.
Index entries for sequences computed from indices in prime factorization.

Cf. A000010, A000026, A000040, A002110, A003958, A005117, A005361, A005867, A006093, A007947, A023900, A048250, A059975, A068997, A081294 (numbers n such that a(n) is odd), A087656, A090624, A152649, A173557, A304117, A304408, A304409, A304411, A304412.

Cf. A001221, A076479.

seq(mul((p-1)*padic[ordp](n, p), p in numtheory[factorset](n)), n=1..100); # Ridouane Oudra, Jun 06 2025

a[n_] := Times @@ ((#[[1]] - 1) #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 75}]
Table[EulerPhi[Last[Select[Divisors[n], SquareFreeQ]]] DivisorSigma[0, n/Last[Select[Divisors[n], SquareFreeQ]]], {n, 75}]

a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); (p-1)*e)} \\ Andrew Howroyd, Jul 24 2018

a(n) = A005361(n)*abs(A023900(n)) = A005361(n)*A173557(n) = A005361(n)*A000010(A007947(n)).
a(p^k) = (p - 1)*k where p is a prime and k > 0.
a(n) = phi(n) if n is a squarefree (A005117), where phi() = A000010.
a(A002110(k)) = A005867(k).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^4/72) * Product_{p prime} (1 - 4/p^2 + 3/p^3 + 1/p^4 - 1/p^5) = 0.2644703894... . - Amiram Eldar, Nov 30 2022
a(n) = (-1)^A001221(n) * (Sum_{d1|n} Sum_{d2|n} mu(d1)*gcd(d1,d2)). - Ridouane Oudra, Jun 06 2025

A345907 Triangle giving the main antidiagonals of the matrices counting integer compositions by length and alternating sum (A345197).

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 2, 2, 1, 1, 0, 0, 4, 3, 1, 1, 0, 0, 3, 6, 4, 1, 1, 0, 0, 6, 9, 8, 5, 1, 1, 0, 0, 0, 18, 18, 10, 6, 1, 1, 0, 0, 0, 10, 36, 30, 12, 7, 1, 1, 0, 0, 0, 20, 40, 60, 45, 14, 8, 1, 1, 0, 0, 0, 0, 80, 100, 90, 63, 16, 9, 1, 1

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

Problem: What are the column sums? They appear to match A239201, but it is not clear why.

			Triangle begins:
   1
   1   1
   0   1   1
   0   1   1   1
   0   2   2   1   1
   0   0   4   3   1   1
   0   0   3   6   4   1   1
   0   0   6   9   8   5   1   1
   0   0   0  18  18  10   6   1   1
   0   0   0  10  36  30  12   7   1   1
   0   0   0  20  40  60  45  14   8   1   1
   0   0   0   0  80 100  90  63  16   9   1   1
   0   0   0   0  35 200 200 126  84  18  10   1   1
   0   0   0   0  70 175 400 350 168 108  20  11   1   1
   0   0   0   0   0 350 525 700 560 216 135  22  12   1   1

Row sums are A163493.
Rows are the antidiagonals of the matrices given by A345197.
The main diagonals of A345197 are A346632, with sums A345908.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A007318, A008549, A025047, A114121, A344610.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{n-k}],k==(n+ats[#])/2-1&]],{k,0,n-1}],{n,0,15}]

A098646 Trace sequence of 3 X 3 Krawtchouk matrix.

Original entry on oeis.org

3, 2, 12, 8, 48, 32, 192, 128, 768, 512, 3072, 2048, 12288, 8192, 49152, 32768, 196608, 131072, 786432, 524288, 3145728, 2097152, 12582912, 8388608, 50331648, 33554432, 201326592, 134217728, 805306368, 536870912, 3221225472, 2147483648

Offset: 0

Paul Barry, Sep 18 2004

Let A=[1,1,1;2,0,-2;1,-1,1], the 3 X 3 Krawtchouk matrix. Then a(n)=trace(A^n).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,4)

Cf. A074535, A098647.

[(-2)^n+2*2^n: n in [0..45]]; // Vincenzo Librandi, Jun 11 2011

G.f.: (3+2*x)/((1+2*x)*(1-2*x)).
a(n) = (-2)^n+2*2^n.
Recurrence: a(n) = 4a(n-2), a(0)=3, a(1)=2. - Ralf Stephan, Jul 17 2013
a(2n+1)=A081294(n+1). a(2n)=A002001(n+1). - R. J. Mathar, Nov 11 2013

A178970 a(n) = 2^(2n-1) mod (2n+1).

Original entry on oeis.org

2, 3, 4, 2, 6, 7, 2, 9, 10, 2, 12, 8, 20, 15, 16, 2, 22, 19, 2, 21, 22, 38, 24, 32, 2, 27, 52, 2, 30, 31, 2, 8, 34, 2, 36, 37, 17, 43, 40, 20, 42, 8, 2, 45, 32, 2, 27, 49, 29, 51, 52, 23, 54, 55, 2, 57, 77, 11, 15, 28, 2, 108, 64, 2, 66, 32, 47, 69, 70, 2

Offset: 1

Juri-Stepan Gerasimov, Jan 03 2011

nonn

			a(1) = 2^(2*1-1) mod (2*1+1) = 2^1 mod 3 = 2.

Cf. A000079.

f[n_] := PowerMod[2, 2 n - 1, 2 n + 1]; Array[f, 70] (* Robert G. Wilson v, Jan 03 2011 *)

a(n) = A081294(n) mod A005408(n).

cp's OEIS Frontend

A053661 For n > 1: if n is present, 2n is not.

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A099142 a(n) = 6^n * T(n, 4/3) where T is the Chebyshev polynomial of the first kind.

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A304411 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)*k_j).

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A333988 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of (1-(k+1)*x) / (1-2*(k+1)*x+((k-1)*x)^2).

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A115716 A divide-and-conquer sequence.

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A134683 Expansion of 1+x*(2+3*x)/(1-4*x^2).

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A304407 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)*k_j).

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A333988 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of (1-(k+1)x) / (1-2(k+1)x+((k-1)x)^2).

A134683 Expansion of 1+x(2+3x)/(1-4*x^2).