A099837 -id:A099837 - cp's OEIS frontend

A054534 Square array giving Ramanujan sum T(n,k) = c_k(n) = Sum_{m=1..k, (m,k)=1} exp(2 Pi i m n / k), read by antidiagonals upwards (n >= 1, k >= 1).

1, 1, -1, 1, 1, -1, 1, -1, -1, 0, 1, 1, 2, -2, -1, 1, -1, -1, 0, -1, 1, 1, 1, -1, 2, -1, -1, -1, 1, -1, 2, 0, -1, -2, -1, 0, 1, 1, -1, -2, 4, -1, -1, 0, 0, 1, -1, -1, 0, -1, 1, -1, 0, 0, 1, 1, 1, 2, 2, -1, 2, -1, -4, -3, -1, -1, 1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, 1, 1, -1, -2, -1, -1, 6, 0, 0, -1, -1, 2, -1, 1, -1, 2, 0, 4, -2, -1, 0, -3, -4, -1, 0, -1, 1

Offset: 1

N. J. A. Sloane, Apr 09 2000

The Ramanujan sum is also known as the von Sterneck arithmetic function. Robert Daublebsky von Sterneck introduced it around 1900. - Petros Hadjicostas, Jul 20 2019

T(n, k) = c_k(n) is the sum of the n-th powers of the k-th primitive roots of unity. - Petros Hadjicostas, Jul 27 2019

			Array T(n,k) (with rows n >= 1 and columns k >= 1) begins as follows:
  1, -1, -1,  0, -1,  1, -1,  0,  0,  1, -1, ...
  1,  1, -1, -2, -1, -1, -1,  0,  0, -1, -1, ...
  1, -1,  2,  0, -1, -2, -1,  0, -3,  1, -1, ...
  1,  1, -1,  2, -1, -1, -1, -4,  0, -1, -1, ...
  1, -1, -1,  0,  4,  1, -1,  0,  0, -4, -1, ...
  1,  1,  2, -2, -1,  2, -1,  0, -3, -1, -1, ...
  1, -1, -1,  0, -1,  1,  6,  0,  0,  1, -1, ...
  ...

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, page 160.
H. Rademacher, Collected Papers of Hans Rademacher, vol. II, MIT Press, 1974, p. 435.
S. Ramanujan, On Certain Trigonometrical Sums and their Applications in the Theory of Numbers, pp. 179-199 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea Publishing 2000.
R. D. von Sterneck, Ein Analogon zur additiven Zahlentheorie, Sitzungsber. Acad. Wiss. Sapientiae Math.-Naturwiss. Kl. 111 (1902), 1567-1601 (Abt. IIa).

Seiichi Manyama, Antidiagonals n = 1..140, flattened
Tom M. Apostol, Arithmetical properties of generalized Ramanujan sums, Pacific J. Math. 41 (1972), 281-293.
Austrian Biographical Encyclopedia from 1815 onwards, Daublebsky von Sterneck, Robert.
Eckford Cohen, A class of arithmetic functions, Proc. Natl. Acad. Sci. USA 41 (1955), 939-944.
A. Elashvili, M. Jibladze, and D. Pataraia, Combinatorics of necklaces and "Hermite reciprocity", J. Algebraic Combin. 10 (1999), 173-188.
M. L. Fredman, A symmetry relationship for a class of partitions, J. Combinatorial Theory Ser. A 18 (1975), 199-202.
Emiliano Gagliardo, Le funzioni simmetriche semplici delle radici n-esime primitive dell'unità, Bollettino dell'Unione Matematica Italiana Serie 3, 8(3) (1953), 269-273.
Otto Hölder, Zur Theorie der Kreisteilungsgleichung K_m(x)=0, Prace mat.-fiz. 43 (1936), 13-23.
Peter H. van der Kamp, On the Fourier transform of the greatest common divisor, Integers 13 (2013), #A24. [See Section 3 for historical remarks.]
C. A. Nicol, On restricted partitions and a generalization of the Euler phi number and the Moebius function, Proc. Natl. Acad. Sci. USA 39(9) (1953), 963-968.
C. A. Nicol and H. S. Vandiver, A von Sterneck arithmetical function and restricted partitions with respect to a modulus, Proc. Natl. Acad. Sci. USA 40(9) (1954), 825-835.
K. G. Ramanathan, Some applications of Ramanujan's trigonometrical sum C_m(n), Proc. Indian Acad. Sci., Sect. A 20 (1944), 62-69.
Srinivasa Ramanujan, On certain trigonometric sums and their applications in the theory of numbers, Trans. Camb. Phil. Soc. 22 (1918), 259-276.
M. V. Subbarao, The Brauer-Rademacher identity, Amer. Math. Monthly 72 (1965), 135-138.
Wikipedia, Ramanujan's sum.
Wikipedia, Robert Daublebsky von Sterneck der Jüngere.
Aurel Wintner, On a statistics of the Ramanujan sums, Amer. J. Math., 64(1) (1942), 106-114.

Cf. A000010, A033999, A054532, A054533, A054535, A062570, A085097, A058384, A085639, A085906, A099837, A100051, A176742, A282634.

nmax = 14; mu[n_Integer] = MoebiusMu[n]; mu[] = 0; t[n, k_] := Total[ #*mu[k/#]& /@ Divisors[n]]; Flatten[ Table[ t[n-k+1, k], {n, 1, nmax}, {k, 1, n}]] (* Jean-François Alcover, Nov 14 2011, after Pari *)
TableForm[Table[t[n, k], {n, 1, 7}, {k, 1, 11}]] (* to print a table like the one in the example - Petros Hadjicostas, Jul 27 2019 *)

{T(n, k) = if( n<1 || k<1, 0, sumdiv( n, d, if( k%d==0, d * moebius(k / d))))} /* Michael Somos, Dec 05 2002 */

{T(n, k) = if( n<1 || k<1, 0, polsym( polcyclo( k), n) [n + 1])} /* Michael Somos, Mar 21 2011 */

/*To get an array like in the example above using Michael Somos' programs:*/
{for (n=1, 20, for (k=1, 40, print1(T(n,k), ","); ); print(); ); } /* Petros Hadjicostas, Jul 27 2019 */

T(n, 1) = c_1(n) = 1. T(n, 2) = c_2(n) = A033999(n). T(n, 3) = c_3(n) = A099837(n) if n>1. T(n, 4) = c_4(n) = A176742(n) if n>1. T(n, 6) = c_6(n) = A100051(n) if n>1. - Michael Somos, Mar 21 2011
T(1, n) = c_n(1) = A008683(n). T(2, n) = c_n(2) = A086831(n). T(3, n) = c_n(3) = A085097(n). T(4, n) = c_n(4) = A085384(n). T(5, n) = c_n(5) = A085639(n). T(6, n) = c_n(6) = A085906(n). - Michael Somos, Mar 21 2011
T(n, n) = T(k * n, n) = A000010(n), T(n, 2*n) = -A062570(n). - Michael Somos, Mar 21 2011
Lambert series and a consequence: Sum_{k >= 1} c_k(n) * z^k / (1 - z^k) = Sum_{s|n} s * z^s and -Sum_{k >= 1} (c_k(n) / k) * log(1 - z^k) = Sum_{s|n} z^s for |z| < 1 (using the principal value of the logarithm). - Petros Hadjicostas, Aug 15 2019

A110162 Riordan array ((1-x)/(1+x), x/(1+x)^2).

Original entry on oeis.org

1, -2, 1, 2, -4, 1, -2, 9, -6, 1, 2, -16, 20, -8, 1, -2, 25, -50, 35, -10, 1, 2, -36, 105, -112, 54, -12, 1, -2, 49, -196, 294, -210, 77, -14, 1, 2, -64, 336, -672, 660, -352, 104, -16, 1, -2, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 2, -100, 825, -2640, 4290, -4004, 2275, -800, 170, -20, 1

Offset: 0

Paul Barry, Jul 14 2005

Inverse of Riordan array A094527. Rows sums are A099837. Diagonal sums are A110164. Product of Riordan array A102587 and inverse binomial transform (1/(1+x), x/(1+x)).
Coefficients of polynomials related to Cartan matrices of types C_n and B_n: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2), with p(x,0) = 1; p(x,1) = 2-x; p(x,2) = x^2-4*x-2. - Roger L. Bagula, Apr 12 2008
From Wolfdieter Lang, Nov 16 2012: (Start)
The alternating row sums are given in A219233.
For n >= 1 the row polynomials in the variable x^2 are R(2*n,x):=2*T(2*n,x/2) with Chebyshev's T-polynomials. See A127672 and also the triangle A127677.
(End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = x/(1 + x)^2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 - 2*x + sqrt(1 - 4*x))/2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

			Triangle T(n,k) begins:
m\k  0    1    2     3     4     5     6    7    8   9 10 ...
0:   1
1:  -2    1
2:   2   -4    1
3:  -2    9   -6     1
4:   2  -16   20    -8     1
5:  -2   25  -50    35   -10     1
6:   2  -36  105  -112    54   -12     1
7:  -2   49 -196   294  -210    77   -14    1
8:   2  -64  336  -672   660  -352   104  -16    1
9:  -2   81 -540  1386 -1782  1287  -546  135  -18   1
10:  2 -100  825 -2640  4290 -4004  2275 -800  170 -20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 16 2012
Row polynomial n=2: P(2,x) = 2 - 4*x + x^2. R(4,x):= 2*T(4,x/2) = 2 - 4*x^2 + x^4. For P and R see a comment above. - _Wolfdieter Lang_, Nov 16 2012.

G. C. Greubel, Rows n=0..100 of triangle, flattened
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv:1405.6315 [math.CO], 2014.

Cf. A128411. See A127677 for an almost identical triangle.

Cf. A136674, A053122.

/* As triangle */ [[(-1)^(n-k)*(Binomial(n+k,n-k) + Binomial(n+k-1,n-k-1)): k in [0..n]]: n in [0.. 12]]; // Vincenzo Librandi, Jun 30 2015

Table[If[n==0 && k==0, 1, (-1)^(n-k)*(Binomial[n+k, n-k] + Binomial[n+k-1, n-k-1])], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 16 2018 *)

{T(n,k) = (-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1))};
for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 16 2018

[[(-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1)) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 16 2018

T(n,k) = (-1)^(n-k)*(C(n+k,n-k) + C(n+k-1,n-k-1)), with T(0,0) = 1. - Paul Barry, Mar 22 2007
From Wolfdieter Lang, Nov 16 2012: (Start)
O.g.f. row polynomials P(n,x) := Sum(T(n,k)*x^k, k=0..n): (1-z^2)/(1+(x-2)*z+z^2) (from the Riordan property).
O.g.f. column No. k: ((1-x)/(1+x))*(x/(1+x)^2)^k, k >= 0.
T(0,0) = 1, T(n,k) = (-1)^(n-k)*(2*n/(n+k))*binomial(n+k,n-k), n>=1, and T(n,k) = 0 if n < k. (From the Chebyshev T-polynomial formula due to Waring's formula.)
(End)
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 29 2013

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 2, 8, 41, 232, 1377, 8399, 52138, 327656, 2077934, 13270633, 85226594, 549837391, 3560702069, 23132584742, 150695482041, 984021596136, 6438849555963, 42208999230224, 277144740254566, 1822379123910857, 11998811140766701, 79095365076843134

Offset: 0

Peter Bala, Mar 07 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence
  {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. For cases, see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^n:
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^1 = 1 + x + O(x^2)
  n = 2: c(x)^2 = 1 + 2*x + 5*x^2 + O(x^3)
  n = 3: c(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + O(x^4)
  n = 4: c(x)^4 = 1 + 4*x + 14*x^2 + 48*x^3 + 165*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 1 = 2, a(2) = 1 + 2 + 5 = 8, a(3) = 1 + 3 + 9 + 28 = 41 and a(4) = 1 + 4 + 14 + 48 + 165 = 232.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                        row sums
  n = 0 |   1                               1
  n = 1 |   1   1                           2
  n = 2 |   5   2    1                      8
  n = 3 |  28   9    3   1                 41
  n = 4 | 165  48   14   4   1            232
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 1, 5, 28, 165, ...] = [x^n] c(x)^n = A025174(n).
Examples of supercongruences:
a(13) - a(1) = 3560702069 - 2 = (3^2)*(13^3)*31*37*157 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11998811140766701 - 41 = (2^2)*5*(7^4)*32213*7756841 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 22794614296746579502 - 1377 = (5^6)*7*53*6491*605796421 == 0 ( mod 5^6 ).

Peter Bala, Notes on A333093

Cf. A000108, A001006, A001764, A005554, A025174, A333090 through A333097, A372214, A372215.

seq(add(n/(n+k)*binomial(n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) -> series(c(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - Sqrt[(1 - 3*x)/(1 + x)]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} n/(n+k)*binomial(n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c(x/(1 + x)) )^n = [x^n] ( (1 + x)*(1 + x*M(x)) )^n, where M(x) = ( 1 - x - sqrt(1 - 2*x - 3*x^2) ) / (2*x^2) is the o.g.f. of the Motzkin numbers A001006.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... = (1/x)*Revert( x/c(x) ) is the o.g.f. of A001764.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3^(3*n + 3/2) / (7 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(n+2*k)*binomial(n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024
D-finite with recurrence 2*n*(2*n-1)*(3991*n -21664)*a(n) +(-1329757*n^3 +9119565*n^2 -18270518*n +10657440)*a(n-1) +10*(947050*n^3 -6943257*n^2 +15944396*n -11260008)*a(n-2) +12*(-787878*n^3 +5778161*n^2 -13283386*n +9383340)*a(n-3) +9*(3*n-10)*(3*n-8)*(100503*n -141587)*a(n-4)=0, n>=5. - R. J. Mathar, Nov 22 2024

A100051 A Chebyshev transform of 1,1,1,...

Original entry on oeis.org

1, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1

Offset: 0

Paul Barry, Oct 31 2004

1, followed by period 6: repeat [1, -1, -2, -1, 1, 2]. - Joerg Arndt, Aug 28 2024
A Chebyshev transform of 1/(1-x): if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).
Transform of 1/(1+x) under the mapping g(x)->((1+x)/(1-x))g(x/(1-x)^2). - Paul Barry, Dec 01 2004
Multiplicative with a(p^e) = -1 if p = 2; -2 if p = 3; 1 otherwise. - David W. Wilson, Jun 10 2005

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Rational Function Multiplicative Coefficients
Index entries for linear recurrences with constant coefficients, signature (1,-1).

Cf. A011655, A057079, A087204, A099837, A099443, A100047, A100048, A100050.

Row sums of array A127677.

CoefficientList[Series[(1 - x^2)/(1 - x + x^2), {x,0,50}], x] (* G. C. Greubel, May 03 2017 *)
LinearRecurrence[{1,-1},{1,1,-1},80] (* Harvey P. Dale, Mar 25 2019 *)

{a(n) = - (n == 0) + [2, 1, -1, -2, -1, 1][n%6 + 1]}; /* Michael Somos, Mar 21 2011 */

From Paul Barry, Dec 01 2004: (Start)
G.f.: (1-x^2)/(1-x+x^2).
a(n) = a(n-1) - a(n-2), n>2.
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)/(n-k).
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*(2n/(n+k))*(-1)^k, n>1. (End)
Moebius transform is length 6 sequence [1, -2, -3, 0, 0, 6].
Euler transform of length 6 sequence [1, -2, -1, 0, 0, 1].
a(n) = a(-n). a(n) = c_6(n) if n>1 where c_k(n) is Ramanujan's sum. - Michael Somos, Mar 21 2011
a(n) = A087204(n), n>0. - R. J. Mathar, Sep 02 2008
a(n) = A057079(n+1), n>0. Dirichlet g.f. zeta(s) *(1-2^(1-s)-3^(1-s)+6^(1-s)). - R. J. Mathar, Apr 11 2011

A230276 Voids left after packing 5-curves coins patterns into fountain of coins with base n.

Original entry on oeis.org

0, 1, 1, 6, 10, 16, 24, 34, 43, 57, 70, 85, 102, 121, 139, 162, 184, 208, 234, 262, 289, 321, 352, 385, 420, 457, 493, 534, 574, 616, 660, 706, 751, 801, 850, 901, 954, 1009, 1063, 1122, 1180, 1240, 1302, 1366, 1429

Offset: 1

Kival Ngaokrajang, Oct 15 2013

Refer to arrangement same as A005169: "A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row". The 5 curves coins patterns consist of a part of each coin circumference and forms a continuous area. There are total 13 distinct patterns. For selected pattern, I would like to call "5C4S" type as it cover 4 coins and symmetry. When packing 5C4S into fountain of coins base n, the total number of 5C4S is A001399, the coins left is A230267 and void left is a(n). See illustration in links.

Kival Ngaokrajang, Illustration of initial terms (V)
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Cf. A008795 (3-curves coins patterns), A074148, A229093, A229154 (4-curves coins patterns), A001399 (5-curves coins patterns), A229593 (6-curves coins patterns).

A099837 := proc(n)
    op(modp(n,3)+1,[2,-1,-1]) ;
end proc:
A230276 := proc(n)
    -A099837(n)/3 + (-48*n+31+18*n^2+9*(-1)^n)/24 ;
end proc:
seq(A230276(n),n=1..40) ; # R. J. Mathar, Feb 28 2018

LinearRecurrence[{1, 1, 0, -1, -1, 1}, {0, 1, 1, 6, 10, 16}, 45] (* Jean-François Alcover, May 05 2023 *)

G.f.: x^2*(x^4 + 3*x^3 + 4*x^2 + 1)/((1-x)*(1-x^2)*(1-x^3)). - Ralf Stephan, Oct 17 2013

a(n) = (9*(-1)^n+18*n^2-48*n)/24 - A099837(n)/3. - R. J. Mathar, Feb 28 2018

A070403 a(n) = 7^n mod 9.

Original entry on oeis.org

1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4, 1, 7, 4

Offset: 0

N. J. A. Sloane, May 12 2002

Also the digital root of 7^n. If we convert this to a repeating decimal 0.174174..., we get the rational number 58/333. - Cino Hilliard, Dec 31 2004
A141722 (1, 25, 121, 505, 2041, 8185) mod 9. Note A141722 = 10*A000975(2n) + A000975(2n+1). - Paul Curtz, Sep 15 2008
Digital root of the powers of any number congruent to 7 mod 9. - Alonso del Arte, Jan 26 2014

Cecil Balmond, Number 9: The Search for the Sigma Code. Munich, New York: Prestel (1998): 203.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 1).

Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 4, A100402; c = 5, A070366; c = 8, A010689.

Cf. A000975, A010888, A049347, A099837, A141722.

[Modexp(7, n, 9): n in [0..110]]; // Bruno Berselli, Mar 22 2016

A070403:=n->4-2*sqrt(3)*sin(2*(n+1)*Pi/3): seq(A070403(n), n=0..100); # Wesley Ivan Hurt, Jun 09 2016

Table[PowerMod[7, n, 9], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

a(n)=7^n%9 \\ Charles R Greathouse IV, Oct 07 2015

[power_mod(7,n,9)for n in range(0,105)] # Zerinvary Lajos, Nov 03 2009

From R. J. Mathar, Feb 23 2009: (Start)
G.f.: (1+7*x+4*x^2)/((1-x)*(1+x+x^2)).
a(n+1) - a(n) = 3*A099837(n+3).
a(n) = 4 - 3*A049347(n). (End)
a(n) = a(n-3) for n>3. - G. C. Greubel, Mar 19 2016
a(n) = 4-2*sqrt(3)*sin((2*n+2)*Pi/3). - Wesley Ivan Hurt, Jun 09 2016
a(n) = A010888(7*a(n-1)). - Stefano Spezia, Mar 20 2025

A115264 Diagonal sums of correlation triangle for floor((n+2)/2).

Original entry on oeis.org

1, 1, 3, 4, 8, 10, 17, 21, 32, 39, 55, 66, 89, 105, 136, 159, 200, 231, 284, 325, 392, 445, 528, 595, 697, 780, 903, 1005, 1152, 1275, 1449, 1596, 1800, 1974, 2211, 2415, 2689, 2926, 3240, 3514, 3872, 4186, 4592, 4950, 5408, 5814, 6328, 6786, 7361

Offset: 0

Paul Barry, Jan 18 2006

Diagonal sums of A115263.
This is associated with the root system F4, and can be described using the additive function on the affine F4 diagram:
2--4--3--2--1
a(n-4) seems to be the number of face-magic cubes or order 2 with magic sum n, which means the sum of the 4 numbers at the 4 corners of each of the 6 faces equals n. (The 8 integers at the corners do not need to be distinct; copies by the 48 operations of rotations and flips are counted only once, cf. A203286, A381589. All 8 integers are positive.). E.g., 1=a(4-4) is the cube with magic sum 4, placing 1 at each corner. 1 =a(5-4) is the number of cubes with magic sum 5 obtained by placing 1 at 6 of the 8 corners but 2 at two corners opposite along a space diagonal. - R. J. Mathar, Mar 11 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1,-2,0,2,1,1,-2,-1,1).

For G2, the corresponding sequence is A001399.
For E6, the corresponding sequence is A164680.
For E7, the corresponding sequence is A210068.
For E8, the corresponding sequence is A045513.
See A210631 for a very similar sequence.

R:=PowerSeriesRing(Integers(), 0); Coefficients(R!( 1/((1-x)*(1-x^2)^2*(1-x^3)*(1-x^4)) )); // G. C. Greubel, Jan 13 2020

seq(coeff(series(1/((1-x)*(1-x^2)^2*(1-x^3)*(1-x^4)), x, n+1), x, n), n = 0..50); # G. C. Greubel, Jan 13 2020

CoefficientList[Series[1/((1-x)*(1-x^2)^2*(1-x^3)*(1-x^4)), {x,0,50}], x] (* G. C. Greubel, Jan 13 2020 *)

A115264(n) := block( A099837(n+3)/27 + A056594(n)/16+(-1)^n*(2*n^2+24*n+63)/256 +(6*n^4 +144*n^3+1194*n^2+3960*n+4267)/6912 )$ /* R. J. Mathar, Mar 19 2012 */

my(x='x+O('x^50)); Vec(1/((1-x)*(1-x^2)^2*(1-x^3)*(1-x^4))) \\ G. C. Greubel, Jan 13 2020

x=PowerSeriesRing(QQ,'x').gen(); 1/((1-x)*(1-x**2)**2*(1-x**3)*(1-x**4))

G.f.: 1/((1-x)*(1-x^2)^2*(1-x^3)*(1-x^4)).
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..n-k} [j<=k]*floor((k-j+2)/2)*[j<=n-2k]*floor((n-2k-j+2)/2).
a(n) = A099837(n+3)/27 + A056594(n)/16 + (-1)^n*(2*n^2 +24*n +63)/256 +(6*n^4 +144*n^3 +1194*n^2 +3960*n +4267)/6912 . - R. J. Mathar, Mar 19 2012

A099838 Expansion of (1-x)^2*(1+x)/(1+x+x^2).

Original entry on oeis.org

1, -2, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0, 3, -3, 0

Offset: 0

Paul Barry, Oct 27 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,-1).

Partial sums are A099837.

Cf. A049347.

[1,-2] cat [3*(n+1 -3*Floor((n+2)/3)): n in [2..110]]; // G. C. Greubel, Apr 21 2023

LinearRecurrence[{-1,-1}, {1,-2,0,3}, 100] (* Jean-François Alcover, Jan 02 2022 *)

[3*(n+1) -9*((n+2)//3) -2*int(n==0) +int(n==1) for n in range(111)] # G. C. Greubel, Apr 21 2023

a(n) = Sum_{k=0..n} (-1)^k*( cos(2*Pi*(n-k)/3) + sin(2*Pi*(n-k)/3)/sqrt(3) )*C(2, k).
a(n) = 2*sqrt(3)*cos((4*n+1)*Pi/6) for n>=2. - Richard Choulet, Apr 23 2009
a(n) = 3*A049347(n) - 2*[n=0] + [n=1]. - G. C. Greubel, Apr 21 2023

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 6, 76, 1101, 16876, 266881, 4305247, 70414133, 1163355884, 19369868385, 324486751951, 5462851474614, 92346622131103, 1566455916243068, 26649562889363259, 454528917186429226, 7769463895152496364, 133064720735632286722, 2282869928179537263601, 39225214245206751480102

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4).
In general, for m > 0 and c(x)^(m*n) is a(n) ~ m * (m+2)^((m+2)*n + 3/2) / (((m+1)*(m+2)+1) * sqrt(2*Pi*n) * (m+1)^((m+1)*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

			n-th order Taylor polynomial of c(x)^(5*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^5 = 1 + 5*x + O(x^2)
  n = 2: c(x)^10 = 1 + 10*x + 65*x^2 + O(x^3)
  n = 3: c(x)^15 = 1 + 15*x + 135*x^2 + 950*x^3 + O(x^4)
  n = 4: c(x)^20 = 1 + 20*x + 230*x^2 + 2000*x^3 + 14625*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 5 = 6, a(2) = 1 + 10 + 65 = 76, a(3) = 1 + 15 + 135 + 950 = 1101 and a(4) = 1 + 20 + 230 + 2000 + 14625 = 16876.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(5*n), n >= 0, in descending powers of x begins
                                                row sums
  n = 0 |     1                                     1
  n = 1 |     5        1                            6
  n = 2 |    65       10      1                    76
  n = 3 |   950      135     15    1             1101
  n = 4 | 14625     2000    230   20    1       16876
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 1566455916243068 - 6 = 2*(13^3)*104701*3404923 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11627033261887689372357353 - 1101 = (2^2)*(7^4)*19*29* 2197177609353575713 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 1034770243516278817426081673131 - 266881 = 2*3*(5^7)*31* 13305359*5351978496238483 == 0 ( mod 5^6 ).

Cf. A000108, A233834, A333090 through A333097.

seq(add(5*n/(5*n+k)*binomial(5*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(5*n), x, 151):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[5*Binomial[7*n-1, n] * HypergeometricPFQ[{1, -6*n, -n}, {1/2 - 7*n/2, 1 - 7*n/2}, 1/4]/6, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 5*n/(5*n+k)*binomial(5*n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^5(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 5*x + 45*x^2 + 500*x^3 + 6200*x^4 + ... = (1/x)*Revert( x/c^5(x) ) is the o.g.f. of A233834.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5 * 7^(7*n + 3/2) / (43 * sqrt(Pi*n) * 2^(6*n + 1) * 3^(6*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 5*n/(5*n+2*k)*binomial(5*n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024

A109044 a(n) = lcm(n,3).

Original entry on oeis.org

0, 3, 6, 3, 12, 15, 6, 21, 24, 9, 30, 33, 12, 39, 42, 15, 48, 51, 18, 57, 60, 21, 66, 69, 24, 75, 78, 27, 84, 87, 30, 93, 96, 33, 102, 105, 36, 111, 114, 39, 120, 123, 42, 129, 132, 45, 138, 141, 48, 147, 150, 51, 156, 159, 54, 165, 168, 57, 174, 177, 60, 183, 186, 63

Offset: 0

Mitch Harris, Jun 18 2005

			G.f. = 3*x + 6*x^2 + 3*x^3 + 12*x^4 + 15*x^5 + 6*x^6 + 21*x^7 + 24*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).
Index entries for sequences related to lcm's.

Cf. A051176, A099837, A109007 (gcd(n,3)), A109042.

[Lcm(n,3): n in [0..63]]; // Bruno Berselli, Mar 11 2011

A109044:=n->lcm(n,3): seq(A109044(n), n=0..100); # Wesley Ivan Hurt, Jul 24 2016

LCM[Range[0, 100], 3] (* Wesley Ivan Hurt, Jul 24 2016 *)

a(n)=lcm(n,3) \\ Charles R Greathouse IV, Oct 07 2015

[lcm(n,3)for n in range(0, 64)] # Zerinvary Lajos, Jun 07 2009

a(n) = 3*n/gcd(n,3) = 3*n/A109007(n).
From Bruno Berselli, Mar 11 2011: (Start)
G.f.: 3*x*(1+2*x+x^2+2*x^3+x^4)/(1-x^3)^2.
a(n) = 3*A051176(n);
a(n) = n*(7-2*A099837(n))/3 for n>0. (End)
From Wesley Ivan Hurt, Jul 24 2016: (Start)
a(n) = 2*a(n-3) - a(n-6) for n>5.
a(n) = 9*n/(5 + 4*cos(2*n*Pi/3)).
If n mod 3 = 0 then 3*floor(n/3), else 3*n. (End)
a(n) = n*(1 + 2*((n^2) mod 3)). - Timothy Hopper, Feb 23 2017
From Michael Somos, Mar 04 2017: (Start)
G.f.: 3 * x / (1 - x)^2 - 6 * x^3 / (1 - x^3)^2. -
a(n) = a(-n) for all n in Z. (End)
Sum_{k=1..n} a(k) ~ (7/6) * n^2. - Amiram Eldar, Nov 26 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(2)/9. - Amiram Eldar, Sep 08 2023

cp's OEIS Frontend

A054534 Square array giving Ramanujan sum T(n,k) = c_k(n) = Sum_{m=1..k, (m,k)=1} exp(2 Pi i m n / k), read by antidiagonals upwards (n >= 1, k >= 1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A110162 Riordan array ((1-x)/(1+x), x/(1+x)^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A100051 A Chebyshev transform of 1,1,1,...

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A230276 Voids left after packing 5-curves coins patterns into fountain of coins with base n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A070403 a(n) = 7^n mod 9.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.