cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A171174 Triangle read by rows in which row n lists A033627(n) together with the first 2n-1 numbers <> 0 of A038608.

Original entry on oeis.org

2, -1, 4, -1, 2, -3, 7, -1, 2, -3, 4, -5, 10, -1, 2, -3, 4, -5, 6, -7, 13, -1, 2, -3, 4, -5, 6, -7, 8, -9, 16, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 19, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 22, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 25, -1, 2, -3, 4
Offset: 1

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Author

Omar E. Pol, Feb 23 2010

Keywords

Comments

Absolute values give A171173.
Note that the partial sums of this sequence gives A171175, a permutation of the natural numbers.

Examples

			Triangle begins:
2, -1,
4, -1,2,-3,
7, -1,2,-3,4,-5,
10,-1,2,-3,4,-5,6,-7,
13,-1,2,-3,4,-5,6,-7,8,-9,
16,-1,2,-3,4,-5,6,-7,8,-9,10,-11,
19,-1,2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,
22,-1,2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15,
25,-1,2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15,16,-17,
		

Crossrefs

A171177 Triangle read by rows in which row n lists 3n-1 together with the first 2n-1 numbers <> 0 of A038608, in reverse order.

Original entry on oeis.org

2, -1, 5, -3, 2, -1, 8, -5, 4, -3, 2, -1, 11, -7, 6, -5, 4, -3, 2, -1, 14, -9, 8, -7, 6, -5, 4, -3, 2, -1, 17, -11, 10, -9, 8, -7, 6, -5, 4, -3, 2, -1, 20, -13, 12, -11, 10, -9, 8, -7, 6, -5, 4, -3, 2, -1, 23, -15, 14, -13, 12, -11, 10, -9, 8, -7, 6, -5, 4, -3, 2, -1, 26, -17, 16, -15
Offset: 1

Views

Author

Omar E. Pol, Feb 23 2010

Keywords

Comments

Absolute values give A171176.
Note that the partial sums of this sequence give A171178, a permutation of the natural numbers.

Examples

			Triangle begins:
2, -1;
5, -3, 2, -1;
8, -5, 4, -3, 2, -1;
11,-7, 6, -5, 4, -3, 2, -1;
14,-9, 8, -7, 6, -5, 4, -3, 2, -1;
17,-11,10,-9, 8, -7, 6, -5, 4, -3, 2, -1;
20,-13,12,-11,10,-9, 8, -7, 6, -5, 4, -3, 2, -1;
23,-15,14,-13,12,-11,10,-9, 8, -7, 6, -5, 4, -3, 2, -1;
26,-17,16,-15,14,-13,12,-11,10,-9, 8, -7, 6, -5, 4, -3, 2, -1;
		

Crossrefs

A001477 The nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 0

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Author

Keywords

Comments

Although this is a list, and lists normally have offset 1, it seems better to make an exception in this case. - N. J. A. Sloane, Mar 13 2010
The subsequence 0,1,2,3,4 gives the known values of n such that 2^(2^n)+1 is a prime (see A019434, the Fermat primes). - N. J. A. Sloane, Jun 16 2010
Also: The identity map, defined on the set of nonnegative integers. The restriction to the positive integers yields the sequence A000027. - M. F. Hasler, Nov 20 2013
The number of partitions of 2n into exactly 2 parts. - Colin Barker, Mar 22 2015
The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 8960 or 168.- Philippe A.J.G. Chevalier, Dec 29 2015
Partial sums give A000217. - Omar E. Pol, Jul 26 2018
First differences are A000012 (the "all 1's" sequence). - M. F. Hasler, May 30 2020
See A061579 for the transposed infinite square matrix, or triangle with rows reversed. - M. F. Hasler, Nov 09 2021
This is the unique sequence (a(n)) that satisfies the inequality a(n+1) > a(a(n)) for all n in N. This simple and surprising result comes from the 6th problem proposed by Bulgaria during the second day of the 19th IMO (1977) in Belgrade (see link and reference). - Bernard Schott, Jan 25 2023

Examples

			Triangular view:
   0
   1   2
   3   4   5
   6   7   8   9
  10  11  12  13  14
  15  16  17  18  19  20
  21  22  23  24  25  26  27
  28  29  30  31  32  33  34  35
  36  37  38  39  40  41  42  43  44
  45  46  47  48  49  50  51  52  53  54
		

References

  • Maurice Protat, Des Olympiades à l'Agrégation, suite vérifiant f(n+1) > f(f(n)), Problème 7, pp. 31-32, Ellipses, Paris 1997.

Crossrefs

Cf. A000027 (n>=1).
Cf. A000012 (first differences).
Partial sums of A057427. - Jeremy Gardiner, Sep 08 2002
Cf. A038608 (alternating signs), A001787 (binomial transform).
Cf. A055112.
Cf. Boustrophedon transforms: A231179, A000737.
Cf. A245422.
Number of orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008586, A008585, A005843, A000217.
When written as an array, the rows/columns are A000217, A000124, A152948, A152950, A145018, A167499, A166136, A167487... and A000096, A034856, A055998, A046691, A052905, A055999... (with appropriate offsets); cf. analogous lists for A000027 in A185787.
Cf. A000290.
Cf. A061579 (transposed matrix / reversed triangle).

Programs

Formula

a(n) = n.
a(0) = 0, a(n) = a(n-1) + 1.
G.f.: x/(1-x)^2.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
When seen as array: T(k, n) = n + (k+n)*(k+n+1)/2. Main diagonal is 2*n*(n+1) (A046092), antidiagonal sums are n*(n+1)*(n+2)/2 (A027480). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: x*e^x. - Franklin T. Adams-Watters, Sep 11 2005
a(0)=0, a(1)=1, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
Alternating partial sums give A001057 = A000217 - 2*(A008794). - Eric Desbiaux, Oct 28 2008
a(n) = 2*A080425(n) + 3*A008611(n-3), n>1. - Eric Desbiaux, Nov 15 2009
a(n) = A007966(n)*A007967(n). - Reinhard Zumkeller, Jun 18 2011
a(n) = Sum_{k>=0} A030308(n,k)*2^k. - Philippe Deléham, Oct 20 2011
a(n) = 2*A028242(n-1) + (-1)^n*A000034(n-1). - R. J. Mathar, Jul 20 2012
a(n+1) = det(C(i+1,j), 1 <= i, j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013
a(n-1) = floor(n/e^(1/n)) for n > 0. - Richard R. Forberg, Jun 22 2013
a(n) = A000027(n) for all n>0.
a(n) = floor(cot(1/(n+1))). - Clark Kimberling, Oct 08 2014
a(0)=0, a(n>0) = 2*z(-1)^[( |z|/z + 3 )/2] + ( |z|/z - 1 )/2 for z = A130472(n>0); a 1 to 1 correspondence between integers and naturals. - Adriano Caroli, Mar 29 2015
G.f. as triangle: x*(1 + (x^2 - 5*x + 2)*y + x*(2*x - 1)*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 22 2025

A008549 Number of ways of choosing at most n-1 items from a set of size 2*n+1.

Original entry on oeis.org

0, 1, 6, 29, 130, 562, 2380, 9949, 41226, 169766, 695860, 2842226, 11576916, 47050564, 190876696, 773201629, 3128164186, 12642301534, 51046844836, 205954642534, 830382690556, 3345997029244, 13475470680616, 54244942336114, 218269673491780, 877940640368572
Offset: 0

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Author

Keywords

Comments

Area under Dyck excursions (paths ending in 0): a(n) is the sum of the areas under all Dyck excursions of length 2*n (nonnegative walks beginning and ending in 0 with jumps -1,+1).
Number of inversions in all 321-avoiding permutations of [n+1]. Example: a(2)=6 because the 321-avoiding permutations of [3], namely 123,132,312,213,231, have 0, 1, 2, 1, 2 inversions, respectively. - Emeric Deutsch, Jul 28 2003
Convolution of A001791 and A000984. - Paul Barry, Feb 16 2005
a(n) = total semilength of "longest Dyck subpath" starting at an upstep U taken over all upsteps in all Dyck paths of semilength n. - David Callan, Jul 25 2008
[1,6,29,130,562,2380,...] is convolution of A001700 with itself. - Philippe Deléham, May 19 2009
From Ran Pan, Feb 04 2016: (Start)
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x to the right. This is related to paired pattern P_2 in Pan and Remmel's link and more details can be found in Section 3.2 in the link.
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) horizontally cross the diagonal y = x. This is related to paired pattern P_3 in Pan and Remmel's link and more details can be found in Section 3.3 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x. This is related to paired pattern P_2 and P_4 in Pan and Remmel's link and more details can be found in Section 4.2 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) cross the diagonal y = x. This is related to paired pattern P_3 and P_4 in Pan and Remmel's link and more details can be found in Section 4.3 in the link. (End)
From Gus Wiseman, Jul 17 2021: (Start)
Also the number of integer compositions of 2*(n+1) with alternating sum < 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(3) = 29 compositions of 8 are:
(1,7) (1,5,2) (1,1,1,5) (1,1,1,4,1) (1,1,1,1,1,3)
(2,6) (1,6,1) (1,1,2,4) (1,2,1,3,1) (1,1,1,2,1,2)
(3,5) (2,5,1) (1,2,1,4) (1,3,1,2,1) (1,1,1,3,1,1)
(1,2,2,3) (1,4,1,1,1) (1,2,1,1,1,2)
(1,3,1,3) (1,2,1,2,1,1)
(1,3,2,2) (1,3,1,1,1,1)
(1,4,1,2)
(1,4,2,1)
(1,5,1,1)
(2,1,1,4)
(2,2,1,3)
(2,3,1,2)
(2,4,1,1)
Also the number of integer compositions of 2*(n+1) with reverse-alternating sum < 0. For a bijection, keep the odd-length compositions and reverse the even-length ones.
Also the number of 2*(n+1)-digit binary numbers with more 0's than 1's. For example, the a(2) = 6 binary numbers are: 100000, 100001, 100010, 100100, 101000, 110000; or in decimal: 32, 33, 34, 36, 40, 48.
(End)

Examples

			a(2) = 6 because there are 6 ways to choose at most 1 item from a set of size 5: You can choose the empty set, or you can choose any of the five one-element sets.
G.f. = x + 6*x^2 + 29*x^3 + 130*x^4 + 562*x^5 + 2380*x^6 + 9949*x^7 + ...
		

References

  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Crossrefs

Odd bisection of A294175 (even is A000346).
For integer compositions of 2*(n+1) with alternating sum k < 0 we have:
- The opposite (k > 0) version is A000302.
- The weak (k <= 0) version is (also) A000302.
- The k = 0 version is A001700 or A088218.
- The reverse-alternating version is also A008549 (this sequence).
- These compositions are ranked by A053754 /\ A345919.
- The complement (k >= 0) is counted by A114121.
- The case of reversed integer partitions is A344743(n+1).
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).
A344610 counts partitions by sum and positive reverse-alternating sum.
A345197 counts compositions by length and alternating sum.

Programs

  • Magma
    [4^n-Binomial(2*n+1, n): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016
    
  • Maple
    A008549:=n->4^n-binomial(2*n+1,n): seq(A008549(n), n=0..30);
  • Mathematica
    Table[4^n-Binomial[2n+1,n],{n,0,30}] (* Harvey P. Dale, May 11 2011 *)
    a[ n_] := If[ n<-4, 0, 4^n - Binomial[2 n + 2, n + 1] / 2] (* Michael Somos, Jan 25 2014 *)
  • PARI
    {a(n)=if(n<0, 0, 4^n - binomial(2*n+1, n))} /* Michael Somos Oct 31 2006 */
    
  • PARI
    {a(n) = if( n<-4, 0, n++; (4^n / 2 - binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */
    
  • Python
    import math
    def C(n,r):
        f=math.factorial
        return f(n)/f(r)/f(n-r)
    def A008549(n):
        return int((4**n)-C(2*n+1,n)) # Indranil Ghosh, Feb 18 2017

Formula

a(n) = 4^n - C(2*n+1, n).
a(n) = Sum_{k=1..n} Catalan(k)*4^(n-k): convolution of Catalan numbers and powers of 4.
G.f.: x*c(x)^2/(1 - 4*x), c(x) = g.f. of Catalan numbers. - Wolfdieter Lang
Note Sum_{k=0..2*n+1} binomial(2*n+1, k) = 2^(2n+1). Therefore, by the symmetry of Pascal's triangle, Sum_{k=0..n} binomial(2*n+1, k) = 2^(2*n) = 4^n. This explains why the following two expressions for a(n) are equal: Sum_{k=0..n-1} binomial(2*n+1, k) = 4^n - binomial(2*n+1, n). - Dan Velleman
G.f.: (2*x^2 - 1 + sqrt(1 - 4*x^2))/(2*(1 + 2*x)*(2*x - 1)*x^3).
a(n) = Sum_{k=0..n} C(2*k, k)*C(2*(n-k), n-k-1). - Paul Barry, Feb 16 2005
Second binomial transform of 2^n - C(n, floor(n/2)) = A045621(n). - Paul Barry, Jan 13 2006
a(n) = Sum_{0 < i <= k < n} binomial(n, k+i)*binomial(n, k-i). - Mircea Merca, Apr 05 2012
D-finite with recurrence (n+1)*a(n) + 2*(-4*n-1)*a(n-1) + 8*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -5. - Michael Somos, Jan 25 2014
Convolution square is A045894. - Michael Somos, Jan 25 2014
HANKEL transform is [0, -1, 2, -3, 4, -5, ...]. - Michael Somos, Jan 25 2014
BINOMIAL transform of [0, 0, 1, 3, 11, 35,...] (A109196) is [0, 0, 1, 6, 29, 130, ...]. - Michael Somos, Jan 25 2014
(n+1) * a(n) = A153338(n+1). - Michael Somos, Jan 25 2014
a(n) = Sum_{m = n+2..2*n+1} binomial(2*n+1,m), n >= 0. - Wolfdieter Lang, May 22 2015
E.g.f.: (exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x))*exp(2*x). - Ilya Gutkovskiy, Aug 30 2016

Extensions

Better description from Dan Velleman (djvelleman(AT)amherst.edu), Dec 01 2000

A374848 Obverse convolution A000045**A000045; see Comments.

Original entry on oeis.org

0, 1, 2, 16, 162, 3600, 147456, 12320100, 2058386904, 701841817600, 488286500625000, 696425232679321600, 2038348954317776486400, 12259459134020160144810000, 151596002479762016373851690400, 3855806813438155578522841251840000
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2024

Keywords

Comments

The obverse convolution of sequences
s = (s(0), s(1), ...) and t = (t(0), t(1), ...)
is introduced here as the sequence s**t given by
s**t(n) = (s(0)+t(n)) * (s(1)+t(n-1)) * ... * (s(n)+t(0)).
Swapping * and + in the representation s(0)*t(n) + s(1)*t(n-1) + ... + s(n)*t(0)
of ordinary convolution yields s**t.
If x is an indeterminate or real (or complex) variable, then for every sequence t of real (or complex) numbers, s**t is a sequence of polynomials p(n) in x, and the zeros of p(n) are the numbers -t(0), -t(1), ..., -t(n).
Following are abbreviations in the guide below for triples (s, t, s**t):
F = (0,1,1,2,3,5,...) = A000045, Fibonacci numbers
L = (2,1,3,4,7,11,...) = A000032, Lucas numbers
P = (2,3,5,7,11,...) = A000040, primes
T = (1,3,6,10,15,...) = A000217, triangular numbers
C = (1,2,6,20,70, ...) = A000984, central binomial coefficients
LW = (1,3,4,6,8,9,...) = A000201, lower Wythoff sequence
UW = (2,5,7,10,13,...) = A001950, upper Wythoff sequence
[ ] = floor
In the guide below, sequences s**t are identified with index numbers Axxxxxx; in some cases, s**t and Axxxxxx differ in one or two initial terms.
Table 1. s = A000012 = (1,1,1,1...) = (1);
t = A000012; 1 s**t = A000079; 2^(n+1)
t = A000027; n s**t = A000142; (n+1)!
t = A000040, P s**t = A054640
t = A000040, P (1/3) s**t = A374852
t = A000079, 2^n s**t = A028361
t = A000079, 2^n (1/3) s**t = A028362
t = A000045, F s**t = A082480
t = A000032, L s**t = A374890
t = A000201, LW s**t = A374860
t = A001950, UW s**t = A374864
t = A005408, 2*n+1 s**t = A000165, 2^n*n!
t = A016777, 3*n+1 s**t = A008544
t = A016789, 3*n+2 s**t = A032031
t = A000142, n! s**t = A217757
t = A000051, 2^n+1 s**t = A139486
t = A000225, 2^n-1 s**t = A006125
t = A032766, [3*n/2] s**t = A111394
t = A034472, 3^n+1 s**t = A153280
t = A024023, 3^n-1 s**t = A047656
t = A000217, T s**t = A128814
t = A000984, C s**t = A374891
t = A279019, n^2-n s**t = A130032
t = A004526, 1+[n/2] s**t = A010551
t = A002264, 1+[n/3] s**t = A264557
t = A002265, 1+[n/4] s**t = A264635
Sequences (c)**L, for c=2..4: A374656 to A374661
Sequences (c)**F, for c=2..6: A374662, A374662, A374982 to A374855
The obverse convolutions listed in Table 1 are, trivially, divisibility sequences. Likewise, if s = (-1,-1,-1,...) instead of s = (1,1,1,...), then s**t is a divisibility sequence for every choice of t; e.g. if s = (-1,-1,-1,...) and t = A279019, then s**t = A130031.
Table 2. s = A000027 = (0,1,2,3,4,5,...) = (n);
t = A000027, n s**t = A007778, n^(n+1)
t = A000290, n^2 s**t = A374881
t = A000040, P s**t = A374853
t = A000045, F s**t = A374857
t = A000032, L s**t = A374858
t = A000079, 2^n s**t = A374859
t = A000201, LW s**t = A374861
t = A005408, 2*n+1 s**t = A000407, (2*n+1)! / n!
t = A016777, 3*n+1 s**t = A113551
t = A016789, 3*n+2 s**t = A374866
t = A000142, n! s**t = A374871
t = A032766, [3*n/2] s**t = A374879
t = A000217, T s**t = A374892
t = A000984, C s**t = A374893
t = A038608, n*(-1)^n s**t = A374894
Table 3. s = A000290 = (0,1,4,9,16,...) = (n^2);
t = A000290, n^2 s**t = A323540
t = A002522, n^2+1 s**t = A374884
t = A000217, T s**t = A374885
t = A000578, n^3 s**t = A374886
t = A000079, 2^n s**t = A374887
t = A000225, 2^n-1 s**t = A374888
t = A005408, 2*n+1 s**t = A374889
t = A000045, F s**t = A374890
Table 4. s = t;
s = t = A000012, 1 s**s = A000079; 2^(n+1)
s = t = A000027, n s**s = A007778, n^(n+1)
s = t = A000290, n^2 s**s = A323540
s = t = A000045, F s**s = this sequence
s = t = A000032, L s**s = A374850
s = t = A000079, 2^n s**s = A369673
s = t = A000244, 3^n s**s = A369674
s = t = A000040, P s**s = A374851
s = t = A000201, LW s**s = A374862
s = t = A005408, 2*n+1 s**s = A062971
s = t = A016777, 3*n+1 s**s = A374877
s = t = A016789, 3*n+2 s**s = A374878
s = t = A032766, [3*n/2] s**s = A374880
s = t = A000217, T s**s = A375050
s = t = A005563, n^2-1 s**s = A375051
s = t = A279019, n^2-n s**s = A375056
s = t = A002398, n^2+n s**s = A375058
s = t = A002061, n^2+n+1 s**s = A375059
If n = 2*k+1, then s**s(n) is a square; specifically,
s**s(n) = ((s(0)+s(n))*(s(1)+s(n-1))*...*(s(k)+s(k+1)))^2.
If n = 2*k, then s**s(n) has the form 2*s(k)*m^2, where m is an integer.
Table 5. Others
s = A000201, LW t = A001950, UW s**t = A374863
s = A000045, F t = A000032, L s**t = A374865
s = A005843, 2*n t = A005408, 2*n+1 s**t = A085528, (2*n+1)^(n+1)
s = A016777, 3*n+1 t = A016789, 3*n+2 s**t = A091482
s = A005408, 2*n+1 t = A000045, F s**t = A374867
s = A005408, 2*n+1 t = A000032, L s**t = A374868
s = A005408, 2*n+1 t = A000079, 2^n s**t = A374869
s = A000027, n t = A000142, n! s**t = A374871
s = A005408, 2*n+1 t = A000142, n! s**t = A374872
s = A000079, 2^n t = A000142, n! s**t = A374874
s = A000142, n! t = A000045, F s**t = A374875
s = A000142, n! t = A000032, L s**t = A374876
s = A005408, 2*n+1 t = A016777, 3*n+1 s**t = A352601
s = A005408, 2*n+1 t = A016789, 3*n+2 s**t = A064352
Table 6. Arrays of coefficients of s(x)**t(x), where s(x) and t(x) are polynomials
s(x) t(x) s(x)**t(x)
n x A132393
n^2 x A269944
x+1 x+1 A038220
x+2 x+2 A038244
x x+3 A038220
nx x+1 A094638
1 x^2+x+1 A336996
n^2 x x+1 A375041
n^2 x 2x+1 A375042
n^2 x x+2 A375043
2^n x x+1 A375044
2^n 2x+1 A375045
2^n x+2 A375046
x+1 F(n) A375047
x+1 x+F(n) A375048
x+F(n) x+F(n) A375049

Examples

			a(0) = 0 + 0 = 0
a(1) = (0+1) * (1+0) = 1
a(2) = (0+1) * (1+1) * (1+0) = 2
a(3) = (0+2) * (1+1) * (1+1) * (2+0) = 16
As noted above, a(2*k+1) is a square for k>=0. The first 5 squares are 1, 16, 3600, 12320100, 701841817600, with corresponding square roots 1, 4, 60, 3510, 837760.
If n = 2*k, then s**s(n) has the form 2*F(k)*m^2, where m is an integer and F(k) is the k-th Fibonacci number; e.g., a(6) = 2*F(3)*(192)^2.
		

Crossrefs

Programs

  • Maple
    a:= n-> (F-> mul(F(n-j)+F(j), j=0..n))(combinat[fibonacci]):
    seq(a(n), n=0..15);  # Alois P. Heinz, Aug 02 2024
  • Mathematica
    s[n_] := Fibonacci[n]; t[n_] := Fibonacci[n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}];
    Table[u[n], {n, 0, 20}]
  • PARI
    a(n)=prod(k=0, n, fibonacci(k) + fibonacci(n-k)) \\ Andrew Howroyd, Jul 31 2024

Formula

a(n) ~ c * phi^(3*n^2/4 + n) / 5^((n+1)/2), where c = QPochhammer(-1, 1/phi^2)^2/2 if n is even and c = phi^(1/4) * QPochhammer(-phi, 1/phi^2)^2 / (phi + 1)^2 if n is odd, and phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 01 2024

A130472 A permutation of the integers: a(n) = (-1)^n * floor( (n+1)/2 ).

Original entry on oeis.org

0, -1, 1, -2, 2, -3, 3, -4, 4, -5, 5, -6, 6, -7, 7, -8, 8, -9, 9, -10, 10, -11, 11, -12, 12, -13, 13, -14, 14, -15, 15, -16, 16, -17, 17, -18, 18, -19, 19, -20, 20, -21, 21, -22, 22, -23, 23, -24, 24, -25, 25, -26, 26, -27, 27, -28, 28, -29, 29, -30, 30, -31, 31, -32, 32
Offset: 0

Views

Author

Clark Kimberling, May 28 2007

Keywords

Comments

Pisano period lengths: 1, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, ... - R. J. Mathar, Aug 10 2012
Partial sums of A038608. - Stanislav Sykora, Nov 27 2013

Crossrefs

Sums of the form Sum_{k=0..n} k^p * q^k: A059841 (p=0,q=-1), this sequence (p=1,q=-1), A089594 (p=2,q=-1), A232599 (p=3,q=-1), A126646 (p=0,q=2), A036799 (p=1,q=2), A036800 (p=2,q=2), A036827 (p=3,q=2), A077925 (p=0,q=-2), A232600 (p=1,q=-2), A232601 (p=2,q=-2), A232602 (p=3,q=-2), A232603 (p=2,q=-1/2), A232604 (p=3,q=-1/2).

Programs

Formula

a(n) = -A001057(n).
a(2n) = n, a(2n+1) = -(n+1).
a(n) = Sum_{k=0..n} k*(-1)^k.
a(n) = -a(n-1) +a(n-2) +a(n-3).
G.f.: -x/( (1-x)*(1+x)^2 ). - R. J. Mathar, Feb 20 2011
a(n) = floor( (n/2)*(-1)^n ). - Wesley Ivan Hurt, Jun 14 2013
a(n) = ceiling( n/2 )*(-1)^n. - Wesley Ivan Hurt, Oct 22 2013
a(n) = ((-1)^n*(2*n+1) - 1)/4. - Adriano Caroli, Mar 28 2015
E.g.f.: (1/4)*(-exp(x) + (1-2*x)*exp(-x) ). - G. C. Greubel, Mar 31 2021

A181983 a(n) = (-1)^(n+1) * n.

Original entry on oeis.org

0, 1, -2, 3, -4, 5, -6, 7, -8, 9, -10, 11, -12, 13, -14, 15, -16, 17, -18, 19, -20, 21, -22, 23, -24, 25, -26, 27, -28, 29, -30, 31, -32, 33, -34, 35, -36, 37, -38, 39, -40, 41, -42, 43, -44, 45, -46, 47, -48, 49, -50, 51, -52, 53, -54, 55, -56, 57, -58, 59
Offset: 0

Views

Author

Michael Somos, Apr 04 2012

Keywords

Comments

This is the Lucas U(-2,1) sequence. - R. J. Mathar, Jan 08 2013
Apparently the Mobius transform of A002129. - R. J. Mathar, Jan 08 2013
For n>0, a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = max(i,j) for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
The sums of the terms of this sequence is the divergent series 1 - 2 + 3 - 4 + ... . Euler summed it to 1/4 which was one of the first examples of summing divergent series. - Michael Somos, Jun 05 2013

Examples

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

References

  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 6.

Crossrefs

Programs

  • Haskell
    a181983 = negate . a038608
    a181983_list = [0, 1] ++ map negate
       (zipWith (+) a181983_list (map (* 2) $ tail a181983_list))
    -- Reinhard Zumkeller, Mar 20 2013
    
  • Magma
    [(-1)^(n+1)*n: n in [0..30]]; // G. C. Greubel, Aug 11 2018
  • Maple
    A181983:=n->-(-1)^n * n; seq(A181983(n), n=0..100); # Wesley Ivan Hurt, Feb 26 2014
  • Mathematica
    a[ n_] := -(-1)^n n;
    a[ n_] := Sign[n] SeriesCoefficient[ x / (1 + x)^2, {x, 0, Abs @n}];
    a[ n_] := Sign[n] (Abs @n)! SeriesCoefficient[ x / Exp[ x], {x, 0, Abs @n}];
    CoefficientList[Series[x/(1+x)^2,{x,0,60}],x] (* or *) LinearRecurrence[{-2,-1},{0,1},60] (* or *) Table[If[OddQ[n],n,-n],{n,0,60}]  (* Harvey P. Dale, Apr 22 2022 *)
  • PARI
    {a(n) = -(-1)^n * n};
    

Formula

G.f.: x / (1 + x)^2.
E.g.f.: x / exp(x).
a(n) = -a(-n) = -(-1)^n * A001477(n) for all n in Z.
a(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = Bernoulli(k) for k = 0, 1, ..., n.
A001787(n) = p(0) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 1, ..., n+1. - Michael Somos, Jun 05 2013
Euler transform of length 2 sequence [-2, 2].
Series reversion of g.f. is A000108(n) (Catalan numbers) with a(0)=0.
Series reversion of e.g.f. is A000169. INVERT transform omitting a(0)=0 is A049347. PSUM transform is A001057. BINOMIAL transform is A154955. - Michael Somos, Jun 05 2013
n * a(n) = A162395(n). - Michael Somos, Jun 05 2013
a(n) = - A038608(n). - Reinhard Zumkeller, Mar 20 2013
a(n+2) = a(n) - 2*(-1)^n. - G. C. Greubel, Aug 11 2018
a(n) = - A274922(n) if n>0. - Michael Somos, Sep 24 2019
From Amiram Eldar, Oct 24 2023: (Start)
Multiplicative with a(2^e) = -2^e, and a(p^e) = p^e for an odd prime p.
Dirichlet g.f.: zeta(s-1) * (1-2^(2-s)). (End)

A076118 a(n) = Sum_{k=n/2..n} k * (-1)^(n-k) * C(k,n-k).

Original entry on oeis.org

0, 1, 1, -1, -3, -2, 2, 5, 3, -3, -7, -4, 4, 9, 5, -5, -11, -6, 6, 13, 7, -7, -15, -8, 8, 17, 9, -9, -19, -10, 10, 21, 11, -11, -23, -12, 12, 25, 13, -13, -27, -14, 14, 29, 15, -15, -31, -16, 16, 33, 17, -17, -35, -18, 18, 37, 19, -19, -39, -20, 20, 41, 21, -21, -43, -22, 22, 45, 23, -23, -47, -24, 24, 49, 25, -25, -51, -26, 26
Offset: 0

Views

Author

Henry Bottomley, Oct 31 2002

Keywords

Comments

Piecewise linear depending on residue modulo 6. Might be described as an inverse Catalan transform of the nonnegative integers.
Number of compositions of n consisting of at most two parts, all congruent to {0,2} mod 3 (offset 1). - Vladeta Jovovic, Mar 10 2005

Examples

			a(10) = -5*1 + 6*15 - 7*35 + 8*28 - 9*9 + 10*1 = -5 + 90 -245 + 224 - 81 + 10 = -7.
		

Crossrefs

Cf. A003881, A038608, A078028, A099254 (partial sums).
See A151842 for a version without signs.

Programs

  • Maple
    A076118:=n->add(k*(-1)^(n-k)*binomial(k,n-k), k=floor(n/2)..n); seq(A076118(n), n=0..50); # Wesley Ivan Hurt, May 08 2014
    f:= gfun:-rectoproc({a(n+4) = 2*a(n+3)-3*a(n+2)+2*a(n+1)-a(n), a(0)=0,a(1)=1,a(2)=1,a(3)=-1}, a(n), remember):
    map(f, [$0..100]); # Robert Israel, Aug 07 2015
  • Mathematica
    Table[Sum[k*(-1)^(n - k)*Binomial[k, n - k], {k, Floor[n/2], n}], {n,
    0, 50}] (* Wesley Ivan Hurt, May 08 2014 *)
  • PARI
    {a(n)=local(k=n%3); n=n\3; (-1)^n*((k>0)+n+(k==1)*n)} /* Michael Somos, Jul 14 2006 */
    
  • PARI
    {a(n)=if(n<0, n=-1-n); polcoeff(x*(1-x)/(1-x+x^2)^2+x*O(x^n),n)} /* Michael Somos, Jul 14 2006 */

Formula

a(3n) = -a(3n-1) = A038608(n).
a(n) = ( 2n*sin((n+1/2)*Pi/3) + sin(n*Pi/3)/sin(Pi/3) )/3.
a(3n) = n*(-1)^n; a(3n+1) = (2n+1)*(-1)^n; a(3n+2) = (n+1)*(-1)^n.
a(n) = Sum{k=0..floor(n/2)} binomial(n-k, k)(-1)^k*(n-k). - Paul Barry, Nov 12 2004
From Michael Somos, Jul 14 2006: (Start)
Euler transform of length 6 sequence [ 1, -2, -2, 0, 0, 2].
G.f.: x(1-x)/(1-x+x^2)^2 = x*(1-x^2)^2*(1-x^3)^2/((1-x)*(1-x^6)^2).
a(-1-n)=a(n). (End)
a(n+4) = 2*a(n+3)-3*a(n+2)+2*a(n+1)-a(n). - Robert Israel, Aug 07 2015
a(n) = A099254(n-1)-A099254(n-2). - R. J. Mathar, Apr 01 2018
Sum_{n>=1} 1/a(n) = Pi/4 (A003881). - Amiram Eldar, May 10 2025

A098493 Triangle T(n,k) read by rows: difference between A098489 and A098490 at triangular rows.

Original entry on oeis.org

1, 0, -1, -1, -1, 1, -1, 1, 2, -1, 0, 3, 0, -3, 1, 1, 2, -5, -2, 4, -1, 1, -2, -7, 6, 5, -5, 1, 0, -5, 0, 15, -5, -9, 6, -1, -1, -3, 12, 9, -25, 1, 14, -7, 1, -1, 3, 15, -18, -29, 35, 7, -20, 8, -1, 0, 7, 0, -42, 14, 63, -42, -20, 27, -9, 1, 1, 4, -22, -24, 85, 14, -112, 42
Offset: 0

Views

Author

Ralf Stephan, Sep 12 2004

Keywords

Comments

Also, coefficients of polynomials that have values in A098495 and A094954.

Examples

			Triangle begins:
   1;
   0, -1;
  -1, -1, 1;
  -1,  1, 2, -1;
   0,  3, 0, -3, 1;
  ...
		

Crossrefs

Columns include A010892, -A076118. Diagonals include A033999, A038608, (-1)^n*A000096. Row sums are in A057077.
Cf. A098494 (diagonal polynomials), A085478, A244419.

Programs

  • Maple
    A098493 := proc (n, k)
    add((-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), j = k..n);
    end proc:
    seq(seq(A098493(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021
  • PARI
    T(n,k)=if(k>n||k<0||n<0,0,if(k>=n-1,(-1)^n*if(k==n,1,-k),if(n==1,0,if(k==0,T(n-1,0)-T(n-2,0),T(n-1,k)-T(n-2,k)-T(n-1,k-1)))))

Formula

T(n, k) = A098489[n(n+1)/2, k] - A098490[n(n+1)/2, k].
Recurrence: T(n, k) = T(n-1, k)-T(n-1, k-1)-T(n-2, k); T(n, k)=0 for n<0, k>n, k<0; T(n, n)=(-1)^n; T(n, n-1)=(-1)^n*(1-n).
G.f.: (1-x)/(1+(y-1)*x+x^2). [Vladeta Jovovic, Dec 14 2009]
From Peter Bala, Jul 13 2021: (Start)
Riordan array ( (1 - x)/(1 - x + x^2), -x/(1 - x + x^2) ).
T(n,k) = (-1)^k * the (n,k)-th entry of Q^(-1)*P = Sum_{j = k..n} (-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), where P denotes Pascal's triangle A007318 and Q denotes triangle A061554 (formed from P by sorting the rows into descending order). (End)
From Peter Bala, Jun 26 2025: (Start)
n-th row polynomial R(n, x) = Sum_{k = 0..n} (-1)^k * binomial(n+k, 2*k) * (1 + x)^k.
R(n, 2*x + 1) = (-1)^n * Dir(n, x), where Dir(n,x) denotes the n-th row polynomial of the triangle A244419.
R(n, -1 - x) = b(n, x), where b(n, x) denotes the n-th row polynomial of the triangle A085478. (End)

A248157 Expansion of (1 - 2*x^2)/(1 + x)^2.

Original entry on oeis.org

1, -2, 1, 0, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 16, -17, 18, -19, 20, -21, 22, -23, 24, -25, 26, -27, 28, -29, 30, -31, 32, -33, 34, -35, 36, -37, 38, -39, 40, -41, 42, -43, 44, -45, 46, -47, 48, -49, 50, -51, 52, -53, 54, -55, 56, -57
Offset: 0

Views

Author

Wolfdieter Lang, Oct 05 2014

Keywords

Comments

First column of Riordan triangle A248156.

Crossrefs

Cf. A038608, A097141, A248156 (row k=0).

Programs

  • Magma
    A248157:= func< n | n eq 0 select 1 else (-1)^n*(3-n) >; // G. C. Greubel, May 16 2025
    
  • Mathematica
    CoefficientList[Series[(1-2x^2)/(1+x)^2,{x,0,60}],x] (* or *) LinearRecurrence[{-2,-1},{1,-2,1},60] (* Harvey P. Dale, Aug 25 2023 *)
  • PARI
    Vec((1 - 2*x^2)/(1 + x)^2 + O(x^80)) \\ Michel Marcus, Oct 11 2014
    
  • SageMath
    def A248157(n): return (-1)^n*(3-n) - 2*int(n==0) # G. C. Greubel, May 16 2025

Formula

O.g.f.: (1 - 2*x^2)/(1 + x)^2 = -2 + 4/(1+x) - 1/(1+x)^2.
a(n) = -2*delta(n,0) + (-1)^n*(3 - n), n >= 0, with Kronecker delta(n,0) = 1 if n=0 else 0.
a(0) = 1, a(n) = -2*a(n-1) - a(n-2), n >= 3, with a(1) = -2, a(2) = 1.
a(n) = A038608(n-3) = A097141(n-1), n>=3.
E.g.f.: (3+x)*exp(-x) - 2. - G. C. Greubel, May 16 2025
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