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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A059841 Period 2: Repeat [1,0]. a(n) = 1 - (n mod 2); Characteristic function of even numbers.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0
Offset: 0

Views

Author

Alford Arnold, Feb 25 2001

Keywords

Comments

When viewed as a triangular array, the row sum values are 0 1 1 1 2 3 3 3 4 5 5 5 6 ... (A004525).
This is the r=0 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Successive binomial transforms of this sequence: A011782, A007051, A007582, A081186, A081187, A081188, A081189, A081190, A060531, A081192.
Characteristic function of even numbers: a(A005843(n))=1, a(A005408(n))=0. - Reinhard Zumkeller, Sep 29 2008
This sequence is the Euler transformation of A185012. - Jason Kimberley, Oct 14 2011
a(n) is the parity of n+1. - Omar E. Pol, Jan 17 2012
Read as partial sequences, we get to A000975. - Jon Perry, Nov 11 2014
Elementary Cellular Automata rule 77 produces this sequence. See Wolfram, Weisstein and Index links below. - Robert Price, Jan 30 2016
Column k = 1 of A051159. - John Keith, Jun 28 2021
When read as a constant: decimal expansion of 10/99, binary expansion of 2/3. - Jason Bard, Aug 25 2025

Examples

			Triangle begins:
  1;
  0, 1;
  0, 1, 0;
  1, 0, 1, 0;
  1, 0, 1, 0, 1;
  0, 1, 0, 1, 0, 1;
  0, 1, 0, 1, 0, 1, 0;
  1, 0, 1, 0, 1, 0, 1, 0;
  1, 0, 1, 0, 1, 0, 1, 0, 1;
  0, 1, 0, 1, 0, 1, 0, 1, 0, 1;
  0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0;
  ...
		

Crossrefs

One's complement of A000035 (essentially the same, but shifted once).
Cf. A033999 (first differences), A008619 (partial sums), A004525, A011782 (binomial transf.), A000975.
Characteristic function of multiples of g: A000007 (g=0), A000012 (g=1), this sequence (g=2), A079978 (g=3), A121262 (g=4), A079998 (g=5), A079979 (g=6), A082784 (g=7).

Programs

  • Haskell
    a059841 n = (1 -) . (`mod` 2)
    a059841_list = cycle [1,0]
    -- Reinhard Zumkeller, May 05 2012, Dec 30 2011
    
  • Magma
    [0^(n mod 2): n in  [0..100]]; // Vincenzo Librandi, Nov 09 2014
    
  • Maple
    seq(1-modp(n,2), n=0..150); # Muniru A Asiru, Apr 05 2018
  • Mathematica
    CoefficientList[Series[1/(1 - x^2), {x, 0, 104}], x] (* or *)
    Array[1/2 + (-1)^#/2 &, 105, 0] (* Michael De Vlieger, Feb 19 2019 *)
    Table[QBinomial[n, 1, -1], {n, 1, 74}] (* John Keith, Jun 28 2021 *)
    PadRight[{},120,{1,0}] (* Harvey P. Dale, Mar 06 2023 *)
  • PARI
    a(n)=(n+1)%2; \\ or 1-n%2 as in NAME.
    
  • PARI
    A059841(n)=!bittest(n,0) \\ M. F. Hasler, Jan 13 2012
    
  • Python
    def A059841(n): return 1 - (n & 1) # Chai Wah Wu, May 25 2022

Formula

a(n) = 1 - A000035(n). - M. F. Hasler, Jan 13 2012
From Paul Barry, Mar 11 2003: (Start)
G.f.: 1/(1-x^2).
E.g.f.: cosh(x).
a(n) = (n+1) mod 2.
a(n) = 1/2 + (-1)^n/2. (End)
Additive with a(p^e) = 1 if p = 2, 0 otherwise.
a(n) = Sum_{k=0..n} (-1)^k*A038137(n, k). - Philippe Deléham, Nov 30 2006
a(n) = Sum_{k=1..n} (-1)^(n-k) for n > 0. - William A. Tedeschi, Aug 05 2011
E.g.f.: cosh(x) = 1 + x^2/(Q(0) - x^2); Q(k) = 8k + 2 + x^2/(1 + (2k + 1)*(2k + 2)/Q(k + 1)); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
E.g.f.: cosh(x) = 1/2*Q(0); Q(k) = 1 + 1/(1 - x^2/(x^2 + (2k + 1)*(2k + 2)/Q(k + 1))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
E.g.f.: cosh(x) = E(0)/(1-x) where E(k) = 1 - x/(1 - x/(x - (2*k+1)*(2*k+2)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Apr 05 2013
For the general case: the characteristic function of numbers that are not multiples of m is a(n) = floor((n-1)/m) - floor(n/m) + 1, m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = A000035(n+1) = A008619(n) - A110654(n). - Wesley Ivan Hurt, Jul 20 2013

Extensions

Better definition from M. F. Hasler, Jan 13 2012
Reinhard Zumkeller's Sep 29 2008 description added as a secondary name by Antti Karttunen, May 03 2022

A008836 Liouville's function lambda(n) = (-1)^k, where k is number of primes dividing n (counted with multiplicity).

Original entry on oeis.org

1, -1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, -1, 1, -1, 1, -1, -1, 1, 1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, 1, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, -1
Offset: 1

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Author

Keywords

Comments

Coons and Borwein: "We give a new proof of Fatou's theorem: if an algebraic function has a power series expansion with bounded integer coefficients, then it must be a rational function. This result is applied to show that for any non-trivial completely multiplicative function from N to {-1,1}, the series sum_{n=1..infinity} f(n)z^n is transcendental over {Z}[z]; in particular, sum_{n=1..infinity} lambda(n)z^n is transcendental, where lambda is Liouville's function. The transcendence of sum_{n=1..infinity} mu(n)z^n is also proved." - Jonathan Vos Post, Jun 11 2008
Coons proves that a(n) is not k-automatic for any k > 2. - Jonathan Vos Post, Oct 22 2008
The Riemann hypothesis is equivalent to the statement that for every fixed epsilon > 0, lim_{n -> infinity} (a(1) + a(2) + ... + a(n))/n^(1/2 + epsilon) = 0 (Borwein et al., theorem 1.2). - Arkadiusz Wesolowski, Oct 08 2013

Examples

			a(4) = 1 because since bigomega(4) = 2 (the prime divisor 2 is counted twice), then (-1)^2 = 1.
a(5) = -1 because 5 is prime and therefore bigomega(5) = 1 and (-1)^1 = -1.
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 37.
  • P. Borwein, S. Choi, B. Rooney and A. Weirathmueller, The Riemann Hypothesis: A Resource for the Aficionado and Virtuoso Alike, Springer, Berlin, 2008, pp. 1-11.
  • H. Gupta, On a table of values of L(n), Proceedings of the Indian Academy of Sciences. Section A, 12 (1940), 407-409.
  • H. Gupta, A table of values of Liouville's function L(n), Research Bulletin of East Panjab University, No. 3 (Feb. 1950), 45-55.
  • P. Ribenboim, Algebraic Numbers, p. 44.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 279.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.3.5 on page 99.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 112.

Crossrefs

Möbius transform of A010052.
Cf. A182448 (Dgf at s=2), A347328 (Dgf at s=3), A347329 (Dgf at s=4).

Programs

  • Haskell
    a008836 = (1 -) . (* 2) . a066829  -- Reinhard Zumkeller, Nov 19 2011
    
  • Maple
    A008836 := n -> (-1)^numtheory[bigomega](n); # Peter Luschny, Sep 15 2011
    with(numtheory): A008836 := proc(n) local i,it,s; it := ifactors(n): s := (-1)^add(it[2][i][2], i=1..nops(it[2])): RETURN(s) end:
  • Mathematica
    Table[LiouvilleLambda[n], {n, 100}] (* Enrique Pérez Herrero, Dec 28 2009 *)
    Table[If[OddQ[PrimeOmega[n]],-1,1],{n,110}] (* Harvey P. Dale, Sep 10 2014 *)
  • PARI
    {a(n) = if( n<1, 0, n=factor(n); (-1)^sum(i=1, matsize(n)[1], n[i,2]))}; /* Michael Somos, Jan 01 2006 */
    
  • PARI
    a(n)=(-1)^bigomega(n) \\ Charles R Greathouse IV, Jan 09 2013
    
  • Python
    from sympy import factorint
    def A008836(n): return -1 if sum(factorint(n).values()) % 2 else 1 # Chai Wah Wu, May 24 2022

Formula

Dirichlet g.f.: zeta(2s)/zeta(s); Dirichlet inverse of A008966.
Sum_{ d divides n } lambda(d) = 1 if n is a square, otherwise 0.
Completely multiplicative with a(p) = -1, p prime.
a(n) = (-1)^A001222(n) = (-1)^bigomega(n). - Jonathan Vos Post, Apr 16 2006
a(n) = A033999(A001222(n)). - Jaroslav Krizek, Sep 28 2009
Sum_{d|n} a(d) *(A000005(d))^2 = a(n) *Sum{d|n} A000005(d). - Vladimir Shevelev, May 22 2010
a(n) = 1 - 2*A066829(n). - Reinhard Zumkeller, Nov 19 2011
a(n) = i^(tau(n^2)-1) where tau(n) is A000005 and i is the imaginary unit. - Anthony Browne, May 11 2016
a(n) = A106400(A156552(n)). - Antti Karttunen, May 30 2017
Recurrence: a(1)=1, n > 1: a(n) = sign(1/2 - Sum_{dMats Granvik, Oct 11 2017
a(n) = Sum_{ d | n } A008683(d)*A010052(n/d). - Jinyuan Wang, Apr 20 2019
a(1) = 1; a(n) = -Sum_{d|n, d < n} mu(n/d)^2 * a(d). - Ilya Gutkovskiy, Mar 10 2021
a(n) = (-1)^A349905(n). - Antti Karttunen, Apr 26 2022
From Ridouane Oudra, Jun 02 2024: (Start)
a(n) = (-1)^A066829(n);
a(n) = (-1)^A063647(n);
a(n) = A101455(A048691(n));
a(n) = sin(tau(n^2)*Pi/2). (End)

A000587 Rao Uppuluri-Carpenter numbers (or complementary Bell numbers): e.g.f. = exp(1 - exp(x)).

Original entry on oeis.org

1, -1, 0, 1, 1, -2, -9, -9, 50, 267, 413, -2180, -17731, -50533, 110176, 1966797, 9938669, 8638718, -278475061, -2540956509, -9816860358, 27172288399, 725503033401, 5592543175252, 15823587507881, -168392610536153, -2848115497132448, -20819319685262839
Offset: 0

Views

Author

Keywords

Comments

Alternating row sums of Stirling2 triangle A048993.
Related to the matrix-exponential of the Pascal-matrix, see A000110 and A011971. - Gottfried Helms, Apr 08 2007
Closely linked to A000110 and especially the contribution there of Jonathan R. Love (japanada11(AT)yahoo.ca), Feb 22 2007, by offering what is a complementary finding.
Number of set partitions of 1..n with an even number of parts, minus the number of such partitions with an odd number of parts. - Franklin T. Adams-Watters, May 04 2010
After -2, the smallest prime is a(36) = -1454252568471818731501051, no others through a(100). What is the first prime >0 in the sequence? - Jonathan Vos Post, Feb 02 2011
a(723) ~ 1.9*10^1265 is almost certainly prime. - D. S. McNeil, Feb 02 2011
Stirling transform of a(n) = [1, -1, 0, 1, 1, ...] is A033999(n) = [1, -1, 1, -1, 1, ...]. - Michael Somos, Mar 28 2012
Negated coefficients in the asymptotic expansion: A005165(n)/n! ~ 1 - 1/n + 1/n^2 + 0/n^3 - 1/n^4 - 1/n^5 + 2/n^6 + 9/n^7 + 9/n^8 - 50/n^9 - 267/n^10 - 413/n^11 + O(1/n^12), starting from the O(1/n) term. - Vladimir Reshetnikov, Nov 09 2016
Named after Venkata Ramamohana Rao Uppuluri and John A. Carpenter of the Mathematics Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee. They are called "Rényi numbers" by Fekete (1999), after the Hungarian mathematician Alfréd Rényi (1921-1970). - Amiram Eldar, Mar 11 2022

Examples

			G.f. = 1 - x + x^3 + x^4 - 2*x^5 - 9*x^6 - 9*x^7 + 50*x^8 + 267*x^9 + 413*x^10 - ...
		

References

  • N. A. Kolokolnikova, Relations between sums of certain special numbers (Russian), in Asymptotic and enumeration problems of combinatorial analysis, pp. 117-124, Krasnojarsk. Gos. Univ., Krasnoyarsk, 1976.
  • Alfréd Rényi, Új modszerek es eredmenyek a kombinatorikus analfzisben. I. MTA III Oszt. Ivozl., Vol. 16 (1966), pp. 7-105.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • M. V. Subbarao and A. Verma, Some remarks on a product expansion. An unexplored partition function, in Symbolic Computation, Number Theory, Special Functions, Physics and Combinatorics (Gainesville, FL, 1999), pp. 267-283, Kluwer, Dordrecht, 2001.

Crossrefs

Cf. A000110, A011971 (base triangle PE), A078937 (PE^2).

Programs

  • Haskell
    a000587 n = a000587_list !! n
    a000587_list = 1 : f a007318_tabl [1] where
       f (bs:bss) xs = y : f bss (y : xs) where y = - sum (zipWith (*) xs bs)
    -- Reinhard Zumkeller, Mar 04 2014
  • Maple
    b:= proc(n, t) option remember; `if`(n=0, 1-2*t,
          add(b(n-j, 1-t)*binomial(n-1, j-1), j=1..n))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..35);  # Alois P. Heinz, Jun 28 2016
  • Mathematica
    Table[ -1 * Sum[ (-1)^( k + 1) StirlingS2[ n, k ], {k, 0, n} ], {n, 0, 40} ]
    With[{nn=30},CoefficientList[Series[Exp[1-Exp[x]],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Nov 04 2011 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ Exp[ 1 - Exp[x]], {x, 0, n}]]; (* Michael Somos, May 27 2014 *)
    a[ n_] := If[ n < 0, 0, With[{m = n + 1}, SeriesCoefficient[ Series[ Nest[ x Factor[ 1 - # /. x -> x / (1 - x)] &, 0, m], {x, 0, m}], {x, 0, m}]]]; (* Michael Somos, May 27 2014 *)
    Table[BellB[n, -1], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k -= b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k; k*(-1)^n, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 09 2019 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( 1 - exp( x + x * O(x^n))), n))}; /* Michael Somos, Mar 14 2011 */
    
  • PARI
    {a(n) = local(A); if( n<0, 0, n++; A = O(x); for( k=1, n, A = x - x * subst(A, x, x / (1 - x))); polcoeff( A, n))}; /* Michael Somos, Mar 14 2011 */
    
  • PARI
    Vec(serlaplace(exp(1 - exp(x+O(x^99))))) /* Joerg Arndt, Apr 01 2011 */
    
  • PARI
    a(n)=round(exp(1)*suminf(k=0,(-1)^k*k^n/k!))
    vector(20,n,a(n-1)) \\ Derek Orr, Sep 19 2014 -- a direct approach
    
  • PARI
    x='x+O('x^66); Vec(serlaplace(exp(1 - exp(x)))) \\ Michel Marcus, Sep 19 2014
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [a(0), a(1), ..., a(n)] for n > 0.
    def A000587_list(n):
        A = [0 for i in range(n)]
        A[n-1] = 1
        R = [1]
        for j in range(0, n):
            A[n-1-j] = -A[n-1]
            for k in range(n-j, n):
                A[k] += A[k-1]
            R.append(A[n-1])
        return R
    # Peter Luschny, Apr 18 2011
    
  • Python
    # Python 3.2 or higher required
    from itertools import accumulate
    A000587, blist, b = [1,-1], [1], -1
    for _ in range(30):
        blist = list(accumulate([b]+blist))
        b = -blist[-1]
        A000587.append(b) # Chai Wah Wu, Sep 19 2014
    
  • Sage
    expnums(26, -1) # Zerinvary Lajos, May 15 2009
    

Formula

a(n) = e*Sum_{k>=0} (-1)^k*k^n/k!. - Benoit Cloitre, Jan 28 2003
E.g.f.: exp(1 - e^x).
a(n) = Sum_{k=0..n} (-1)^k S2(n, k), where S2(i, j) are the Stirling numbers of second kind A008277.
G.f.: (x/(1-x))*A(x/(1-x)) = 1 - A(x); the binomial transform equals the negative of the sequence shifted one place left. - Paul D. Hanna, Dec 08 2003
With different signs: g.f.: Sum_{k>=0} x^k/Product_{L=1..k} (1 + L*x).
Recurrence: a(n) = -Sum_{i=0..n-1} a(i)*C(n-1, i). - Ralf Stephan, Feb 24 2005
Let P be the lower-triangular Pascal-matrix, PE = exp(P-I) a matrix-exponential in exact integer arithmetic (or PE = lim exp(P)/exp(1) as limit of the exponential); then a(n) = PE^-1 [n,1]. - Gottfried Helms, Apr 08 2007
Take the series 0^n/0! - 1^n/1! + 2^n/2! - 3^n/3! + 4^n/4! + ... If n=0 then the result will be 1/e, where e = 2.718281828... If n=1, the result will be -1/e. If n=2, the result will be 0 (i.e., 0/e). As we continue for higher natural number values of n sequence for the Roa Uppuluri-Carpenter numbers is generated in the numerator, i.e., 1/e, -1/e, 0/e, 1/e, 1/e, -2/e, -9/e, -9/e, 50/e, 267/e, ... . - Peter Collins (pcolins(AT)eircom.net), Jun 04 2007
The sequence (-1)^n*a(n), with general term Sum_{k=0..n} (-1)^(n-k)*S2(n, k), has e.g.f. exp(1-exp(-x)). It also has Hankel transform (-1)^C(n+1,2)*A000178(n) and binomial transform A109747. - Paul Barry, Mar 31 2008
G.f.: 1 / (1 + x / (1 - x / (1 + x / (1 - 2*x / (1 + x / (1 - 3*x / (1 + x / ...))))))). - Michael Somos, May 12 2012
From Sergei N. Gladkovskii, Sep 28 2012 to Feb 07 2014: (Start)
Continued fractions:
G.f.: -1/U(0) where U(k) = x*k - 1 - x + x^2*(k+1)/U(k+1).
G.f.: 1/(U(0)+x) where U(k) = 1 + x - x*(k+1)/(1 + x/U(k+1)).
G.f.: 1+x/G(0) where G(k) = x*k - 1 + x^2*(k+1)/G(k+1).
G.f.: (1 - G(0))/(x+1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x+1/G(k+1) )).
G.f.: 1 + x/(G(0)-x) where G(k) = x*k + 2*x - 1 - x*(x*k+x-1)/G(k+1).
G.f.: G(0)/(1+x), where G(k) = 1-x^2*(k+1)/(x^2*(k+1)+(x*k-1-x)*(x*k-1)/G(k+1)).
(End)
a(n) = B_n(-1), where B_n(x) is n-th Bell polynomial. - Vladimir Reshetnikov, Oct 20 2015
From Mélika Tebni, May 20 2022: (Start)
a(n) = Sum_{k=0..n} (-1)^k*Bell(k)*A129062(n, k).
a(n) = Sum_{k=0..n} (-1)^k*k!*A130191(n, k). (End)

A005449 Second pentagonal numbers: a(n) = n*(3*n + 1)/2.

Original entry on oeis.org

0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480
Offset: 0

Views

Author

Keywords

Comments

Number of edges in the join of the complete graph and the cycle graph, both of order n, K_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Also number of cards to build an n-tier house of cards. - Martin Wohlgemuth, Aug 11 2002
The modular form Delta(q) = q*Product_{n>=1} (1-q^n)^24 = q*(1 + Sum_{n>=1} (-1)^n*(q^(n*(3*n-1)/2)+q^(n*(3*n+1)/2)))^24 = q*(1 + Sum_{n>=1} A033999(n)*(q^A000326(n)+q^a(n)))^24. - Jonathan Vos Post, Mar 15 2006
Row sums of triangle A134403.
Bisection of A001318. - Omar E. Pol, Aug 22 2011
Sequence found by reading the line from 0 in the direction 0, 7, ... and the line from 2 in the direction 2, 15, ... in the square spiral whose vertices are the generalized pentagonal numbers, A001318. - Omar E. Pol, Sep 08 2011
A general formula for the n-th second k-gonal number is given by T(n, k) = n*((k-2)*n+k-4)/2, n>=0, k>=5. - Omar E. Pol, Aug 04 2012
Partial sums give A006002. - Denis Borris, Jan 07 2013
A002260 is the following array A read by antidiagonals:
0, 1, 2, 3, 4, 5, ...
0, 1, 2, 3, 4, 5, ...
0, 1, 2, 3, 4, 5, ...
0, 1, 2, 3, 4, 5, ...
0, 1, 2, 3, 4, 5, ...
0, 1, 2, 3, 4, 5, ...
and a(n) is the hook sum: Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
From Klaus Purath, May 13 2021: (Start)
This sequence and A000326 provide all integers m such that 24*m + 1 is a square. The union of the two sequences is A001318.
If A is a sequence satisfying the recurrence t(n) = 3*t(n-1) - 2*t(n-2) with the initial values either A(0) = 1, A(1) = n + 2 or A(0) = -1, A(1) = n - 1, then a(n) = (A(i)^2 - A(i-1)*A(i+1))/2^i + n^2 for i>0. (End)
a(n+1) is the number of Dyck paths of size (3,3n+2), i.e., the number of NE lattice paths from (0,0) to (3,3n+2) which stay above the line connecting these points. - Harry Richman, Jul 13 2021
Binomial transform of [0, 2, 3, 0, 0, 0, ...], being a(n) = 2*binomial(n,1) + 3*binomial(n,2). a(3) = 15 = [0, 2, 3, 0] dot [1, 3, 3, 1] = [0 + 6 + 9 + 0]. - Gary W. Adamson, Dec 17 2022
a(n) is the sum of longest side length of all nondegenerate integer-sided triangles with shortest side length n and middle side length (n + 1), n > 0. - Torlach Rush, Feb 04 2024

Examples

			From _Omar E. Pol_, Aug 22 2011: (Start)
Illustration of initial terms:
                                               O
                                             O O
                                 O         O O O
                               O O       O O O O
                     O       O O O     O O O O O
                   O O     O O O O     O O O O O
           O     O O O     O O O O     O O O O O
         O O     O O O     O O O O     O O O O O
    O    O O     O O O     O O O O     O O O O O
    O    O O     O O O     O O O O     O O O O O
    -    ---     -----     -------     ---------
    2     7        15         26           40
(End)
		

References

  • Henri Cohen, A Course in Computational Algebraic Number Theory, vol. 138 of Graduate Texts in Mathematics, Springer-Verlag, 2000.

Crossrefs

Cf. A016789 (first differences), A006002 (partial sums).
The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, this sequence, A045943, A115067, A140090, A140091, A059845, A140672-A140675, A151542.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=3).
Cf. numbers of the form n*((2*k+1)*n+1)/2 listed in A022289 (this sequence is the case k=1).

Programs

Formula

a(n) = A110449(n, 1) for n>0.
G.f.: x*(2+x)/(1-x)^3. E.g.f.: exp(x)*(2*x + 3*x^2/2). a(n) = n*(3*n + 1)/2. a(-n) = A000326(n). - Michael Somos, Jul 18 2003
a(n) = A001844(n) - A000217(n+1) = A101164(n+2,2) for n>0. - Reinhard Zumkeller, Dec 03 2004
a(n) = Sum_{j=1..n} n+j. - Zerinvary Lajos, Sep 12 2006
a(n) = A126890(n,n). - Reinhard Zumkeller, Dec 30 2006
a(n) = 2*C(3*n,4)/C(3*n,2), n>=1. - Zerinvary Lajos, Jan 02 2007
a(n) = A000217(n) + A000290(n). - Zak Seidov, Apr 06 2008
a(n) = a(n-1) + 3*n - 1 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010
a(n) = A129267(n+5,n). - Philippe Deléham, Dec 21 2011
a(n) = 2*A000217(n) + A000217(n-1). - Philippe Deléham, Mar 25 2013
a(n) = A130518(3*n+1). - Philippe Deléham, Mar 26 2013
a(n) = (12/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Vladimir Kruchinin, Jun 04 2013
a(n) = floor(n/(1-exp(-2/(3*n)))) for n>0. - Richard R. Forberg, Jun 22 2013
a(n) = Sum_{i=1..n} (3*i - 1) for n >= 1. - Wesley Ivan Hurt, Oct 11 2013 [Corrected by Rémi Guillaume, Oct 24 2024]
a(n) = (A000292(6*n+k+1)-A000292(k))/(6*n+1) - A000217(3*n+k+1), for any k >= 0. - Manfred Arens, Apr 26 2015
Sum_{n>=1} 1/a(n) = 6 - Pi/sqrt(3) - 3*log(3) = 0.89036376976145307522... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(2*n) - A000217(n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/sqrt(3) + 4*log(2) - 6. - Amiram Eldar, Jan 18 2021
From Klaus Purath, May 13 2021: (Start)
Partial sums of A016789 for n > 0.
a(n) = 3*n^2 - A000326(n).
a(n) = A000326(n) + n.
a(n) = A002378(n) + A000217(n-1) for n >= 1. [Corrected by Rémi Guillaume, Aug 14 2024] (End)
From Klaus Purath, Jul 14 2021: (Start)
b^2 = 24*a(n) + 1 we get by b^2 = (a(n+1) - a(n-1))^2 = (a(2*n)/n)^2.
a(2*n) = n*(a(n+1) - a(n-1)), n > 0.
a(2*n+1) = n*(a(n+1) - a(n)). (End)
A generalization of Lajos' formula, dated Sep 12 2006, follows. Let SP(k,n) = the n-th second k-gonal number. Then SP(2k+1,n) = Sum_{j=1..n} (k-1)*n+j+k-2. - Charlie Marion, Jul 13 2024
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * binomial(k, 2) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
For integer m, (6*m + 1)^2*a(n) + a(m) = a((6*m+1)*n + m). - Peter Bala, Jan 09 2025

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A025147 Number of partitions of n into distinct parts >= 2.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 8, 10, 12, 15, 17, 21, 25, 29, 35, 41, 48, 56, 66, 76, 89, 103, 119, 137, 159, 181, 209, 239, 273, 312, 356, 404, 460, 522, 591, 669, 757, 853, 963, 1085, 1219, 1371, 1539, 1725, 1933, 2164, 2418, 2702, 3016, 3362, 3746, 4171, 4637, 5155
Offset: 0

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Author

Keywords

Comments

From R. J. Mathar, Jul 31 2008: (Start)
These "partitions of n into distinct parts >= k" and "partitions of n into distinct parts, the least being k-1" come in pairs of similar, almost shifted but not identical, sequences:
The distinction in the definitions is that "distinct parts >= k" sets a lower bound to all parts, whereas "the least being ..." means that the lower limit must be attained by one of the parts. (End)
From N. J. A. Sloane, Sep 28 2008: (Start)
Generating functions and Maple programs for the sequences in the first and second columns of the above list are respectively:
For A025147, A025148, etc.:
f:=proc(k) product(1+x^j, j=k..100): series(%,x,100): seriestolist(%); end;
For A096765, A096749, etc.:
g:=proc(k) x^(k-1)*product(1+x^j, j=k..100): series(%,x,100): seriestolist(%); end; (End)
Also number of partitions of n+1 into distinct parts, the least being 1.
Number of different sums from 1+[1,3]+[1,4]+...+[1,n]. - Jon Perry, Jan 01 2004
Also number of partitions of n such that if k is the largest part, then all parts from 1 to k occur, k occurring at least twice. Example: a(7)=3 because we have [2,2,2,1],[2,2,1,1,1] and [1,1,1,1,1,1,1]. - Emeric Deutsch, Apr 09 2006
Also number of partitions of n+1 such that if k is the largest part, then all parts from 1 to k occur, k occurring exactly once. Example: a(7)=3 because we have [3,2,2,1],[3,2,1,1,1] and [2,1,1,1,1,1,1] (there is a simple bijection with the partitions defined before). - Emeric Deutsch, Apr 09 2006
Also number of partitions of n+1 into distinct parts where the number of parts is itself a part. - Reinhard Zumkeller, Nov 04 2007
Partial sums give A038348 (observed by Jonathan Vos Post, proved by several correspondents).
Trivially, number of partitions of n into distinct parts (as ascending lists) such that the first part is not 1, the second not 2, the third not 3, etc., see example. - Joerg Arndt, Jun 10 2013
Convolution with A033999 gives A270144 (apart from the offset). - R. J. Mathar, Jun 18 2016

Examples

			a(7) = 3, from {{3, 4}, {2, 5}, {7}}
From _Joerg Arndt_, Jun 10 2013: (Start)
There are a(17) = 21 partitions of 17 into distinct parts >=2:
01:  [ 2 3 4 8 ]
02:  [ 2 3 5 7 ]
03:  [ 2 3 12 ]
04:  [ 2 4 5 6 ]
05:  [ 2 4 11 ]
06:  [ 2 5 10 ]
07:  [ 2 6 9 ]
08:  [ 2 7 8 ]
09:  [ 2 15 ]
10:  [ 3 4 10 ]
11:  [ 3 5 9 ]
12:  [ 3 6 8 ]
13:  [ 3 14 ]
14:  [ 4 5 8 ]
15:  [ 4 6 7 ]
16:  [ 4 13 ]
17:  [ 5 12 ]
18:  [ 6 11 ]
19:  [ 7 10 ]
20:  [ 8 9 ]
21:  [ 17 ]
(End)
		

References

  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education, Vol. 31, No. 1, pp. 24-28, Winter 1997. MathEduc Database (Zentralblatt MATH, 1997c.01891).
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17. Zentralblatt MATH, Zbl 1071.05501.

Crossrefs

Programs

  • Haskell
    a025147 = p 2 where
       p _ 0 = 1
       p k m = if m < k then 0 else p (k + 1) (m - k) + p (k + 1) m
    -- Reinhard Zumkeller, Dec 28 2011
    
  • Maple
    g:=product(1+x^j,j=2..65): gser:=series(g,x=0,62): seq(coeff(gser,x,n),n=0..57); # Emeric Deutsch, Apr 09 2006
    with(combstruct):ZL := {L = PowerSet(Sequence(Z,card>=2)) },unlabeled:seq(count([L,ZL],size=i),i=0..57); # Zerinvary Lajos, Mar 09 2007
  • Mathematica
    CoefficientList[Series[Product[1+q^n, {n, 2, 60}], {q, 0, 60}], q]
    FoldList[ PartitionsQ[ #2+1 ]-#1&, 0, Range[ 64 ] ]
    (* also *)
    d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 && Min[#] >= 2 &]; Table[d[n], {n, 12}] (* strict partitions, parts >= 2 *)
    Table[Length[d[n]], {n, 40}] (* A025147 for n >= 1 *)
    (* Clark Kimberling, Mar 07 2014 *)
    p[, 0] = 1; p[k, m_] := p[k, m] = If[m < k, 0, p[k+1, m-k] + p[k+1, m]]; Table[p[2, m], {m, 0, 59}] (* Jean-François Alcover, Apr 17 2014, after Reinhard Zumkeller *)
  • PARI
    a(n)=if(n,my(v=partitions(n));sum(i=1,#v,v[i][1]>1&&v[i]==vecsort(v[i],,8)),1) \\ Charles R Greathouse IV, Nov 20 2012

Formula

G.f.: Product_{k>=2} (1+x^k).
a(n) = A000009(n)-a(n-1) = Sum_{0<=k<=n} (-1)^k*A000009(n-k). - Henry Bottomley, May 09 2002
a(n)=t(n, 1), where t(n, k)=1+Sum_{i>j>k and i+j=n} t(i, j), 2<=k<=n. - Reinhard Zumkeller, Jan 01 2003
G.f.: 1 + Sum_{k=1..infinity} (x^(k*(k+3)/2) / Product_{j=1..k} (1-x^j)). - Emeric Deutsch, Apr 09 2006
The previous g.f. is a special case of the g.f. for partitions into distinct parts >= L, Sum_{n>=0} ( x^(n*(n+2*L-1)/2) / Product_{k=1..n} (1-x^k) ). - Joerg Arndt, Mar 24 2011
G.f.: Sum_{n>=1} ( x^(n*(n+1)/2-1) / Product_{k=1..n-1} (1-x^k) ), a special case of the g.f. for partitions into distinct parts >= L, Sum_{n>=L-1} ( x^(n*(n+1)/2-L*(L-1)/2) / Product_{k=1..n-(L-1)} (1-x^k) ). - Joerg Arndt, Mar 27 2011
a(n) = Sum_{1A060016(n-k+1,k-1), for n>0. - Reinhard Zumkeller, Nov 04 2007
a(n) = A096765(n+1). - R. J. Mathar, Jul 31 2008
From Vaclav Kotesovec, Aug 16 2015: (Start)
a(n) ~ 1/2 * A000009(n).
a(n) ~ exp(Pi*sqrt(n/3)) / (8*3^(1/4)*n^(3/4)).
(End)

Extensions

Corrected and extended by Dean Hickerson, Oct 10 2001

A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

Original entry on oeis.org

0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600
Offset: 0

Views

Author

Keywords

Comments

a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series with alternate signs of gamma matrix with determinant 1!.
a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.
A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.
a(n) is even with respect to signs of power of A.
See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)
Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009
This is a divisibility sequence.
a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013
This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
Differences between successive golden rectangle numbers A001654. - Jonathan Sondow, Nov 05 2015
a(n+1) is the number of 2 X n matrices that can be obtained from a 2 X n matrix by moving each element to an adjacent position, horizontally or vertically. This is because F(n+1) is the number of domino tilings of that matrix, therefore with a checkerboard coloring and two domino tilings we can move the black element of each domino of the first tiling to the white element of the same domino and similarly move the white element of each domino of the second tiling to the black element of the same domino. - Fabio Visonà, May 04 2022
In general, squaring the terms of a second-order linear recurrence with signature (c,d) will result in a third-order linear recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Jan 05 2023

Examples

			G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...
		

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.
  • Ross Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Richard P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

Crossrefs

Bisection of A006498 and A074677. First differences of A001654.
Second row of array A103323.
Half of A175395.

Programs

  • GAP
    List([0..30], n -> Fibonacci(n)^2); # G. C. Greubel, Dec 10 2018
    
  • Haskell
    a007598 = (^ 2) . a000045  -- Reinhard Zumkeller, Sep 01 2013
    
  • Magma
    [Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Apr 14 2011
    
  • Maple
    with(combinat): seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007
  • Mathematica
    f[n_] := Fibonacci[n]^2; Array[f, 4!, 0] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2009 *)
    LinearRecurrence[{2,2,-1},{0,1,1},41] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    {a(n) = fibonacci(n)^2};
    
  • PARI
    concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Altug Alkan, Nov 06 2015
    
  • Python
    from sympy import fibonacci
    def A007598(n): return fibonacci(n)**2 # Chai Wah Wu, Apr 14 2025
  • Sage
    [(fibonacci(n))^2 for n in range(0, 28)]# Zerinvary Lajos, May 15 2009
    
  • Sage
    [fibonacci(n)^2 for n in range(30)] # G. C. Greubel, Dec 10 2018
    

Formula

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.
a(-n) = a(n) for all n in Z.
a(n) = A080097(n-2) + 1.
L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} a(n)/n*x^n; special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011
a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g., a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004
a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004
a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).
Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)*A005248(n+1) for n >= 1 [P. Stanica]. Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n*A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006
a(n) = (A005248(n) - 2*(-1)^n)/5. - R. J. Mathar, Sep 12 2010
a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - Gary Detlefs, Dec 13 2010
a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - Gary Detlefs, Dec 20 2010
a(n) = Fibonacci(2*n-2) + a(n-2). - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011
a(n) = (((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n + (3-sqrt(5))*((3-sqrt(5))/2)^n + 4*(-1)^n)/10. - Tim Monahan, Jul 17 2011
E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012
Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - Gary W. Adamson, Mar 06 2012
a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013
a(n) + a(n-1) = A001519(n), n > 0. - R. J. Mathar, Mar 19 2014
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].
a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = (F(n+2)*F(n+3) - L(n)*L(n+1))/3 for F = A000045 and L = A000032. - J. M. Bergot, Jun 02 2014
0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014
(F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n) = A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015
a(n) = F(n)*F(n+1) - F(n-1)*F(n). - Jonathan Sondow, Nov 05 2015
For n>2, a(n) = F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015
a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015
a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016
a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - J. M. Bergot, May 25 2017
4*a(n) = L(n+1)*L(n-1) - F(n+2)*F(n-2), where L = A000032. - Bruno Berselli, Sep 27 2017
a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - Federico Provvedi, Dec 10 2018
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.
Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) = (10 - 3*sqrt(5))/18.
Conjecture: Sum_{n >= 1, n != 2*k+1} 1/(a(n) + (-1)^n*a(2*k+1)) = 1/a(4*k+2) for k = 0,1,2,.... (End)
Sum_{n>=1} 1/a(n) = A105393. - Amiram Eldar, Oct 22 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi (A001622) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A057077 Periodic sequence 1,1,-1,-1; expansion of (1+x)/(1+x^2).

Original entry on oeis.org

1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 04 2000

Keywords

Comments

Abscissa of the image produced after n alternating reflections of (1,1) over the x and y axes respectively. Similarly, the ordinate of the image produced after n alternating reflections of (1,1) over the y and x axes respectively. - Wesley Ivan Hurt, Jul 06 2013

Crossrefs

Programs

Formula

G.f.: (1+x)/(1+x^2).
a(n) = S(n, 0) + S(n-1, 0) = S(2*n, sqrt(2)); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 0)=A056594.
a(n) = (-1)^binomial(n,2) = (-1)^floor(n/2) = 1/2*((n+2) mod 4 - n mod 4). For fixed r = 0,1,2,..., it appears that (-1)^binomial(n,2^r) gives a periodic sequence of period 2^(r+1), the period consisting of a block of 2^r plus ones followed by a block of 2^r minus ones. See A033999 (r = 0), A143621 (r = 2) and A143622 (r = 3). Define E(k) = sum {n = 0..inf} a(n)*n^k/n! for k = 0,1,2,... . Then E(0) = cos(1) + sin(1), E(1) = cos(1) - sin(1) and E(k) is an integral linear combination of E(0) and E(1) (a Dobinski-type relation). Precisely, E(k) = A121867(k) * E(0) - A121868(k) * E(1). See A143623 and A143624 for the decimal expansions of E(0) and E(1) respectively. For a fixed value of r, similar relations hold between the values of the sums E_r(k) := Sum_{n>=0} (-1)^floor(n/r)*n^k/n!, k = 0,1,2,... . For particular cases see A000587 (r = 1) and A143628 (r = 3). - Peter Bala, Aug 28 2008
Sum_{k>=0} a(k)/(k+1) = Sum_{k>=0} 1/((a(k)*(k+1))) = log(2)/2 + Pi/4. - Jaume Oliver Lafont, Apr 30 2010
a(n) = (-1)^A180969(1,n), where the first index in A180969(.,.) is the row index. - Adriano Caroli, Nov 18 2010
a(n) = (-1)^((2*n+(-1)^n-1)/4) = i^((n-1)*n), with i=sqrt(-1). - Bruno Berselli, Dec 27 2010 - Aug 26 2011
Non-simple continued fraction expansion of (3+sqrt(5))/2 = A104457. - R. J. Mathar, Mar 08 2012
E.g.f.: cos(x)*(1 + tan(x)). - Arkadiusz Wesolowski, Aug 31 2012
From Ricardo Soares Vieira, Oct 15 2019: (Start)
E.g.f.: sin(x) + cos(x) = sqrt(2)*sin(x + Pi/4).
a(n) = sqrt(2)*(d^n/dx^n) sin(x)|_x=Pi/4, i.e., a(n) equals sqrt(2) times the n-th derivative of sin(x) evaluated at x=Pi/4. (End)
a(n) = 4*floor(n/4) - 2*floor(n/2) + 1. - Ridouane Oudra, Mar 23 2024

A053632 Irregular triangle read by rows giving coefficients in expansion of Product_{k=1..n} (1 + x^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 7, 7, 6, 5, 5, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4
Offset: 0

Views

Author

N. J. A. Sloane, Mar 22 2000

Keywords

Comments

Or, triangle T(n,k) read by rows, giving number of subsets of {1,2,...,n} with sum k. - Roger CUCULIERE (cuculier(AT)imaginet.fr), Nov 19 2000
Row n consists of A000124(n) terms. These are also the successive vectors (their nonzero elements) when one starts with the infinite vector (of zeros) with 1 inserted somewhere and then shifts it one step (right or left) and adds to the original, then shifts the result two steps and adds, three steps and adds, etc. - Antti Karttunen, Feb 13 2002
T(n,k) = number of partitions of k into distinct parts <= n. Triangle of distribution of Wilcoxon's signed rank statistic. - Mitch Harris, Mar 23 2006
T(n,k) = number of binary words of length n in which the sum of the positions of the 0's is k. Example: T(4,5)=2 because we have 0110 (sum of the positions of the 0's is 1+4=5) and 1001 (sum of the positions of the 0's is 2+3=5). - Emeric Deutsch, Jul 23 2006
A fair coin is flipped n times. You receive i dollars for a "success" on the i-th flip, 1<=i<=n. T(n,k)/2^n is the probability that you will receive exactly k dollars. Your expectation is n(n+1)/4 dollars. - Geoffrey Critzer, May 16 2010
From Gus Wiseman, Jan 02 2023: (Start)
With offset 1, also the number of integer compositions of n whose partial sums add up to k for k = n..n(n+1)/2. For example, row n = 6 counts the following compositions:
6 15 24 33 42 51 141 231 321 411 1311 2211 3111 12111 21111 111111
114 123 132 222 312 1131 1221 2121 11121 11211
213 1113 1122 1212 2112 1111
(End)

Examples

			Triangle begins:
  1;
  1, 1;
  1, 1, 1, 1;
  1, 1, 1, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 3, 2, 2, 1, 1, 1;
  ...
Row n = 4 counts the following binary words, where k = sum of positions of zeros:
  1111  0111  1011  0011  0101  0110  0001  0010  0100  1000  0000
                    1101  1110  1001  1010  1100
Row n = 5 counts the following strict partitions of k with all parts <= n (0 is the empty partition):
  0  1  2  3  4  5  42  43  53  54  532  542  543  5431 5432 54321
           21 31 32 51  52  431 432 541  5321 5421
                 41 321 421 521 531 4321
		

References

  • A. V. Yurkin, New binomial and new view on light theory, (book), 2013, 78 pages, no publisher listed.

Crossrefs

Rows reduced modulo 2 and interpreted as binary numbers: A068052, A068053. Rows converge towards A000009.
Row sums give A000079.
Cf. A285101 (multiplicative encoding of each row), A285103 (number of odd terms on row n), A285105 (number of even terms).
Row lengths are A000124.
A reciprocal version is (A033999, A219977, A291983, A291984, A291985, ...).
A negative version is A231599.
A version for partitions is A358194, reversed partitions A264034.

Programs

  • Maple
    with(gfun,seriestolist); map(op,[seq(seriestolist(series(mul(1+(z^i), i=1..n),z,binomial(n+1,2)+1)), n=0..10)]); # Antti Karttunen, Feb 13 2002
    # second Maple program:
    g:= proc(n) g(n):= `if`(n=0, 1, expand(g(n-1)*(1+x^n))) end:
    T:= n-> seq(coeff(g(n), x, k), k=0..degree(g(n))):
    seq(T(n), n=0..10);  # Alois P. Heinz, Nov 19 2012
  • Mathematica
    Table[CoefficientList[ Series[Product[(1 + t^i), {i, 1, n}], {t, 0, 100}], t], {n, 0, 8}] // Grid (* Geoffrey Critzer, May 16 2010 *)

Formula

From Mitch Harris, Mar 23 2006: (Start)
T(n,k) = T(n-1, k) + T(n-1, k-n), T(0,0)=1, T(0,k) = 0, T(n,k) = 0 if k < 0 or k > (n+1 choose 2).
G.f.: (1+x)*(1+x^2)*...*(1+x^n). (End)
Sum_{k>=0} k * T(n,k) = A001788(n). - Alois P. Heinz, Feb 09 2017
max_{k>=0} T(n,k) = A025591(n). - Alois P. Heinz, Jan 20 2023

A131689 Triangle of numbers T(n,k) = k!*Stirling2(n,k) = A000142(k)*A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
(1,1,1) (1,2,2) (1,2,3)
(2,1,2) (1,3,2)
(2,2,1) (2,1,3)
(1,1,2) (2,3,1)
(1,2,1) (3,1,2)
(2,1,1) (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

Examples

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)
		

Crossrefs

Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) + T(n-1, k))
    end
    for n in 0:7
        println([T(n, k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
    
  • Maple
    A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
    # Alternatively:
    A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017
  • Mathematica
    t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
    T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    
  • SageMath
    @cached_function
    def F(n): # Fubini polynomial
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

Formula

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018
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