A047336 -id:A047336 - cp's OEIS frontend

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

1, 5, 10, 17, 26, 36, 49, 62, 79, 95, 116, 135, 160, 182, 211, 236, 269, 297, 334, 365, 406, 440, 485, 522, 571, 611, 664, 707, 764, 810, 871, 920, 985, 1037, 1106, 1161, 1234, 1292, 1369, 1430, 1511, 1575, 1660, 1727, 1816, 1886, 1979, 2052, 2149, 2225, 2326

Offset: 0

Bruno Berselli, Oct 20 2011 - based on remarks and sequences by Omar E. Pol

For an origin of this sequence, see the numerical spiral illustrated in the Links section.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A195020 (vertices of the numerical spiral in link).

Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A195021, A195023-A195030, A195320, A198017 [incomplete list].

[(14*n*(n+3)+(2*n-5)*(-1)^n+21)/16: n in [0..50]];

Table[(14*n*(n + 3) + (2*n - 5)*(-1)^n + 21)/16, {n, 0, 50}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{1,5,10,17,26},60] (* Harvey P. Dale, Jun 19 2020 *)

for(n=0, 50, print1((14*n*(n+3)+(2*n-5)*(-1)^n+21)/16", "));

O.g.f.: (1 + 4*x + 3*x^2 - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: (1/16)*((21 + 56*x + 14*x^2)*exp(x) - (5 + 2*x)*exp(-x)). - G. C. Greubel, Aug 19 2017
a(n) = A195020(n) + n + 1.
a(n)   - a(-n-1)  = A047336(n+1).
a(n+1) - a(-n)    = A113804(n+1).
a(n+2) - a(n)     = A047385(n+3).
a(n+4) - a(n)     = A113803(n+4).
a(2*n) + a(2*n-1) = A069127(n+1).
a(2*n) - a(2*n-1) = A016813(n).
a(2*n+1) - a(2*n) = A016777(n+1).
a(n+2) + 2*a(n+1) + a(n) = A024966(n+2).

A198442 Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (1,1,0) or (1,0,0).

Original entry on oeis.org

0, 0, 2, 3, 6, 8, 12, 15, 20, 24, 30, 35, 42, 48, 56, 63, 72, 80, 90, 99, 110, 120, 132, 143, 156, 168, 182, 195, 210, 224, 240, 255, 272, 288, 306, 323, 342, 360, 380, 399, 420, 440, 462, 483, 506, 528, 552, 575, 600, 624, 650, 675, 702, 728, 756, 783, 812

Offset: 1

Paul Weisenhorn, Oct 25 2011

If the sequence ends with (1,1,0) Abel wins; if it ends with (1,0,0) Kain wins.
Abel(n) = A002620(n-1) = (2*n*(n - 2) + 1 - (-1)^n)/8.
Kain(n) = A004526(n-1) = floor((n - 1)/2).
Win probability for Abel = sum(Abel(n)/2^n) = 2/3.
Win probability for Kain = sum(Kain(n)/2^n) = 1/3.
Mean length of the game = sum(n*a(n)/2^n) = 16/3.
Essentially the same as A035106. - R. J. Mathar, Oct 27 2011
The sequence 2*a(n) is denoted as chi(n) by McKee (1994) and is the degree of the division polynomial f_n as a polynomial in x. He notes that "If x is given weight 1, a is given weight 2, and b is given weight 3, then all the terms in f_n(a, b, x) have weight chi(n)". - Michael Somos, Jan 09 2015
In Duistermaat (2010), at the end of section 11.2 The Elliptic Billiard, on page 492 the number of k-periodic fibers counted with multiplicities of the QRT root is given by equation (11.2.8) as "1/4 k^2 + 3{k/2}(1 - {k/2}) - 1 = n^2 - 1 when k = 2n, n^2 + n when k = 2n+1, for every integer k." - Michael Somos, Mar 14 2023

			For n = 6 the a(6) = 8 solutions are (0,0,0,1,1,0), (0,1,0,1,1,0),(0,0,1,1,1,0), (1,0,1,1,1,0), (0,1,1,1,1,0),(1,1,1,1,1,0) for Abel and
  (0,0,0,1,0,0), (0,1,0,1,0,0) for Kain.
G.f. = 2*x^3 + 3*x^4 + 6*x^5 + 8*x^6 + 12*x^7 + 15*x^8 + 20*x^9 + ...

J. J. Duistermaat, Discrete Integrable Systems, 2010, Springer Science+Business Media.
A. Engel, Wahrscheinlichkeitsrechnung und Statistik, Band 2, Klett, 1978, pages 25-26.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
J. McKee, Computing division polynomials, Math. Comp. 63 (1994), 767-771. MR1248973 (95a:11110)
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000004, A002620, A004526, A004652, A005843, A008585, A008586, A023443, A028242, A047221, A047336, A052928, A242477, A265611.

Cf. A035106, A079524, A238377.

[(2*n^2-5-3*(-1)^n)/8: n in [1..60]]; // Vincenzo Librandi, Oct 28 2011

for n from 1 by 2 to 99 do
  a(n):=(n^2-1)/4:
  a(n+1):=(n+1)^2/4-1:
end do:
seq(a(n),n=1..100);

a[ n_] := Quotient[ n^2 - 1, 4]; (* Michael Somos, Jan 09 2015 *)

a(n)=([1,1,0,0,0,0;0,0,1,1,0,0;0,1,0,0,1,0;0,0,0,1,1,0;0,0,0,0,0,2;0,0,0,0,0,2]^n)[1,5] \\ Charles R Greathouse IV, Oct 26 2011

{a(n) = (n^2 - 1) \ 4}; /* Michael Somos, Jan 09 2015 */

sub a {
    my ($t, $n) = (0, shift);
    for (0..((1<<$n)-1)) {
        my $str = substr unpack("B32", pack("N", $_)), -$n;
        $t++ if ($str =~ /1.0$/ and not $str =~ /1.0./);
    }
    return $t
} # Charles R Greathouse IV, Oct 26 2011

def A198442():
    yield 0
    x, y = 0, 2
    while True:
       yield x
       x, y = x + y, x//y + 1
a = A198442(); print([next(a) for i in range(57)]) # Peter Luschny, Dec 22 2015

a(n) = (2*n^2 - 5 - 3*(-1)^n)/8.
a(2*n) = n^2 - 1; a(2*n+1) = n*(n + 1).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) with n>=4.
G.f.: x^3*(2 - x)/((1 + x)*(1 - x)^3). - R. J. Mathar, Oct 27 2011
a(n) = a(-n) for all n in Z. a(0) = -1. - Michael Somos, Jan 09 2015
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-1 - a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 09 2015
1 = a(n) - a(n+1) - a(n+2) + a(n+3), 2 = a(n) - 2*a(n+2) + a(n+4) for all n in Z. - Michael Somos, Jan 09 2015
a(n) = A002620(n+2) - A052928(n+2) for n >= 1. (Note A265611(n) = A002620(n+1) + A052928(n+1) for n >= 1.) - Peter Luschny, Dec 22 2015
a(n+1) = A110654(n)^2 + A110654(n)*(2 - (n mod 2)), n >= 0. - Fred Daniel Kline, Jun 08 2016
a(n) = A004526(n)*A004526(n+3). - Fred Daniel Kline, Aug 04 2016
a(n) = floor((n^2 - 1)/4). - Bruno Berselli, Mar 15 2021

a(12) inserted by Charles R Greathouse IV, Oct 26 2011

A175886 Numbers that are congruent to {1, 12} mod 13.

Original entry on oeis.org

1, 12, 14, 25, 27, 38, 40, 51, 53, 64, 66, 77, 79, 90, 92, 103, 105, 116, 118, 129, 131, 142, 144, 155, 157, 168, 170, 181, 183, 194, 196, 207, 209, 220, 222, 233, 235, 246, 248, 259, 261, 272, 274, 285, 287, 298, 300, 311, 313, 324, 326, 337, 339, 350

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 13).

Bruno Berselli, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A091998, A113801, A005408, A047209, A007310, A047336, A047522, A056020, A090771, A175885, A175887.

Cf. A195045 (partial sums).

a175886 n = a175886_list !! (n-1)
a175886_list = 1 : 12 : map (+ 13) a175886_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..350] | n mod 13 in [1, 12]]; // Bruno Berselli, Feb 29 2012

[(26*n+9*(-1)^n-13)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 350], MemberQ[{1, 12}, Mod[#, 13]]&] (* Bruno Berselli, Feb 29 2012 *)
CoefficientList[Series[(1 + 11 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{1,1,-1},{1,12,14},60] (* Harvey P. Dale, Oct 23 2015 *)

a(n)=(26*n+9*(-1)^n-13)/4 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(1+11*x+x^2)/((1+x)*(1-x)^2).
a(n) = (26*n+9*(-1)^n-13)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = a(n-2)+13.
a(n) = 13*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n>1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/13)*cot(Pi/13). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((26*x - 13)*exp(x) + 9*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/13).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/13)*cosec(Pi/13). (End)

A175887 Numbers that are congruent to {1, 14} mod 15.

Original entry on oeis.org

1, 14, 16, 29, 31, 44, 46, 59, 61, 74, 76, 89, 91, 104, 106, 119, 121, 134, 136, 149, 151, 164, 166, 179, 181, 194, 196, 209, 211, 224, 226, 239, 241, 254, 256, 269, 271, 284, 286, 299, 301, 314, 316, 329, 331, 344, 346, 359, 361, 374, 376, 389, 391, 404

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in this case, a(n)^2-1==0 (mod 15).

Bruno Berselli, Table of n, a(n) for n = 1..10000.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A019693, A019976, A113801, A175886, A091998, A175885, A090771, A056020, A047522, A047336, A007310, A047209, A005408, A001651.

Cf. A132240 (primes).

a175887 n = a175887_list !! (n-1)
a175887_list = 1 : 14 : map (+ 15) a175887_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..450] | n mod 15 in [1,14]];

[(30*n+11*(-1)^n-15)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 450], MemberQ[{1,14}, Mod[#, 15]]&]
CoefficientList[Series[(1 + 13 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)

a(n)=(30*n+11*(-1)^n-15)/4 \\ Charles R Greathouse IV, Sep 28 2015

G.f.: x*(1+13*x+x^2)/((1+x)*(1-x)^2).
a(n) = (30*n+11*(-1)^n-15)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = 15*A000217(n-1) -2*sum(a(i), i=1..n-1) +1 for n>1.
a(n) = A047209(A047225(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/15)*cot(Pi/15) = A019693 * A019976 / 10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((30*x - 15)*exp(x) + 11*exp(-x))/4. - David Lovler, Sep 05 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/15)*cosec(Pi/15).
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*cos(Pi/15). (End)

A195041 Concentric heptagonal numbers.

Original entry on oeis.org

0, 1, 7, 15, 28, 43, 63, 85, 112, 141, 175, 211, 252, 295, 343, 393, 448, 505, 567, 631, 700, 771, 847, 925, 1008, 1093, 1183, 1275, 1372, 1471, 1575, 1681, 1792, 1905, 2023, 2143, 2268, 2395, 2527, 2661, 2800, 2941, 3087, 3235, 3388, 3543

Offset: 0

Omar E. Pol, Sep 27 2011

A033582 and A069127 interleaved.

Partial sums of A047336. - Reinhard Zumkeller, Jan 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Column 7 of A195040.

Cf. A032527, A032528, A033582, A047336, A069099, A069127, A077221, A195042.

a195041 n = a195041_list !! n
a195041_list = scanl (+) 0 a047336_list
-- Reinhard Zumkeller, Jan 07 2012

[7*n^2/4+3*((-1)^n-1)/8: n in [0..50]]; // Vincenzo Librandi, Sep 29 2011

CoefficientList[Series[-((x (1+5 x+x^2))/((-1+x)^3 (1+x))),{x,0,80}],x] (* or *) LinearRecurrence[{2,0,-2,1},{0,1,7,15},80] (* Harvey P. Dale, Jan 18 2021 *)

a(n)=7*n^2\4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 7*n^2/4 + 3*((-1)^n - 1)/8.
From R. J. Mathar, Sep 28 2011: (Start)
G.f.: -x*(1+5*x+x^2) / ( (1+x)*(x-1)^3 ).
a(n) + a(n+1) = A069099(n+1). (End)
a(n) = n^2 + floor(3*n^2/4). - Bruno Berselli, Aug 08 2013
Sum_{n>=1} 1/a(n) = Pi^2/42 + tan(sqrt(3/7)*Pi/2)*Pi/sqrt(21). - Amiram Eldar, Jan 16 2023

A045472 Primes congruent to {1, 6} mod 7.

Original entry on oeis.org

13, 29, 41, 43, 71, 83, 97, 113, 127, 139, 167, 181, 197, 211, 223, 239, 251, 281, 293, 307, 337, 349, 379, 419, 421, 433, 449, 461, 463, 491, 503, 547, 587, 601, 617, 631, 643, 659, 673, 701, 727, 743, 757, 769

Offset: 1

N. J. A. Sloane

Primes p such that p^4 = 1 mod 210. - Gary Detlefs, Dec 29 2011

Primes in A047336, also in A113801. - Reinhard Zumkeller, Jan 07 2012

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A042989 (complement), A010051.

a045472 n = a045472_list !! (n-1)
a045472_list = [x | x <- a047336_list, a010051 x == 1]
-- Reinhard Zumkeller, Jan 07 2012

[ p: p in PrimesUpTo(1000) | p mod 7 in {1,6} ]; // Vincenzo Librandi, Aug 13 2012

Select[Prime[Range[200]],MemberQ[{1,6},Mod[#,7]]&] (* Vincenzo Librandi, Aug 13 2012 *)

select(p->abs(centerlift(Mod(p,7)))==1, primes(100)) \\ Charles R Greathouse IV, Mar 17 2022

a(n) ~ 3n log n. - Charles R Greathouse IV, Mar 17 2022

A186439 Numbers whose squares end in three identical digits.

Original entry on oeis.org

38, 100, 200, 300, 400, 462, 500, 538, 600, 700, 800, 900, 962, 1000, 1038, 1100, 1200, 1300, 1400, 1462, 1500, 1538, 1600, 1700, 1800, 1900, 1962, 2000, 2038, 2100, 2200, 2300, 2400, 2462, 2500, 2538, 2600, 2700, 2800, 2900, 2962, 3000, 3038, 3100, 3200, 3300, 3400, 3462

Offset: 1

Michel Lagneau, Feb 21 2011

The three ending digits of a(n)^2 are 000 or 444.

n is in the sequence iff either n == 0 mod 100 or n == (+/-)38 mod 500. - Robert Israel, Jul 03 2014

			462 is in the sequence because 462^2 = 213444.

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).

Cf. A016742 (even squares), A186438.
Cf. A039685, A047336.
Cf. A346678.

with(numtheory):T:=array(1..10):for p from 1 to 10000 do:n:=p^2:l:=length(n):n0:=n:for
  m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :T[m]:=u:od:if T[1]=T[2]
  and T[1]=T[3] then printf(`%d, `,p):else fi:od:
# second Maple program:
a:= proc(n) local m, r;
      r:= 1+ irem(n-1, 7, 'm');
      [38, 100, 200, 300, 400, 462, 500][r] +500*m
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 24 2011

Select[Range[11,10000],Mod[PowerMod[#,2,1000],111]==0&] (* Zak Seidov, Feb 23 2011 *)

for(n=11,10000,if((n^2%1000)%111==0,print1(n", "))) \\ Zak Seidov, Feb 23 2011

Vec(2*x*(19*x^2 +12*x +19)*(x^4 +x^3 +x^2 +x +1)/((x -1)^2*(x^6 +x^5 +x^4 +x^3 +x^2 +x +1)) + O(x^100)) \\ Colin Barker, Jul 03 2014

def ok(n): s = str(n*n); return len(s) > 2 and s[-1] == s[-2] == s[-3]
print(list(filter(ok, range(3463)))) # Michael S. Branicky, Jul 29 2021

a(A047336(n)) = A039685(n). - Bruno Berselli, Feb 22 2011
a(n) = a(n-7) + 500 for n > 7. - Zak Seidov and Bruno Berselli, Feb 23 2011
G.f.: 2*x*(19*x^2 +12*x +19)*(x^4 +x^3 +x^2 +x +1) / ((x -1)^2*(x^6 +x^5 +x^4 +x^3 +x^2 +x +1)). - Colin Barker, Jul 03 2014

A047283 Numbers that are congruent to {0, 1, 3, 6} mod 7.

Original entry on oeis.org

0, 1, 3, 6, 7, 8, 10, 13, 14, 15, 17, 20, 21, 22, 24, 27, 28, 29, 31, 34, 35, 36, 38, 41, 42, 43, 45, 48, 49, 50, 52, 55, 56, 57, 59, 62, 63, 64, 66, 69, 70, 71, 73, 76, 77, 78, 80, 83, 84, 85, 87, 90, 91, 92, 94, 97, 98, 99, 101, 104, 105, 106, 108, 111

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047336, A047355.

[n : n in [0..150] | n mod 7 in [0, 1, 3, 6]]; // Wesley Ivan Hurt, May 22 2016

A047283:=n->(14*n-15+I^(2*n)+(3+I)*I^(-n)+(3-I)*I^n)/8: seq(A047283(n), n=1..100); # Wesley Ivan Hurt, May 22 2016

Select[Range[0,100], MemberQ[{0,1,3,6}, Mod[#,7]]&] (* or *) LinearRecurrence[{1,0,0,1,-1}, {0,1,3,6,7}, 60] (* Harvey P. Dale, Mar 09 2012 *)

G.f.: x^2*(1+2*x+3*x^2+x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - Harvey P. Dale, Mar 09 2012
From Wesley Ivan Hurt, May 22 2016: (Start)
a(n) = (14n-15+i^(2n)+(3+i)*i^(-n)+(3-i)*i^n)/8 where i=sqrt(-1).
a(2n) = A047336(n), a(2n-1) = A047355(n). (End)

More terms from Wesley Ivan Hurt, May 22 2016

A047322 Numbers that are congruent to {0, 1, 5, 6} mod 7.

Original entry on oeis.org

0, 1, 5, 6, 7, 8, 12, 13, 14, 15, 19, 20, 21, 22, 26, 27, 28, 29, 33, 34, 35, 36, 40, 41, 42, 43, 47, 48, 49, 50, 54, 55, 56, 57, 61, 62, 63, 64, 68, 69, 70, 71, 75, 76, 77, 78, 82, 83, 84, 85, 89, 90, 91, 92, 96, 97, 98, 99, 103, 104, 105, 106, 110, 111

Offset: 1

N. J. A. Sloane

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

Cf. A030308, A047336, A047382.

[n : n in [0..150] | n mod 7 in [0, 1, 5, 6]]; // Wesley Ivan Hurt, May 23 2016

A047322:=n->(14*n-11-3*I^(2*n)+(3-3*I)*I^(-n)+(3+3*I)*I^n)/8: seq(A047322(n), n=1..100); # Wesley Ivan Hurt, May 23 2016

Table[(14n-11-3*I^(2n)+(3-3*I)*I^(-n)+(3+3*I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, May 23 2016 *)
LinearRecurrence[{1, 0, 0, 1, -1}, {0, 1, 5, 6, 7}, 60] (* Vincenzo Librandi, May 24 2016 *)

a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=5, b(k)=7*2^(k-2) for k>1. - Philippe Deléham, Oct 19 2011
G.f.: x^2*(1+4*x+x^2+x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 03 2011
From Wesley Ivan Hurt, May 23 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14n-11-3*I^(2n)+(3-3*I)*I^(-n)+(3+3*I)*I^n)/8 where I=sqrt(-1).
a(2n) = A047336(n), a(2n-1) = A047382(n). (End)
E.g.f.: (4 - 3*sin(x) + 3*cos(x) + (7*x - 4)*sinh(x) + 7*(x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 24 2016

More terms from Wesley Ivan Hurt, May 23 2016

A323674 Square array, read by antidiagonals, of the positive integers 6cd +-c +-d = (6c +- 1)d +- c. Alternate rows (or columns) are numbers that differ by c from multiples of 6c - 1 or 6c + 1.

Original entry on oeis.org

4, 6, 6, 9, 8, 9, 11, 13, 13, 11, 14, 15, 20, 15, 14, 16, 20, 24, 24, 20, 16, 19, 22, 31, 28, 31, 22, 19, 21, 27, 35, 37, 37, 35, 27, 21, 24, 29, 42, 41, 48, 41, 42, 29, 24, 26, 34, 46, 50, 54, 54, 50, 46, 34, 26, 29, 36, 53, 54, 65, 60, 65, 54, 53, 36, 29, 31, 41, 57, 63, 71, 73, 73, 71, 63, 57, 41, 31

Offset: 1

Sally Myers Moite, Jan 23 2019

This sequence without duplicates is A067611, which is the complement of A002822, the positive integers x for which 6x - 1 and 6x + 1 are twin primes.

			Square array begins:
   4,   6,   9,  11,  14,  16,  19,  21,  24,  26, ...
   6,   8,  13,  15,  20,  22,  27,  29,  34,  36, ...
   9,  13,  20,  24,  31,  35,  42,  46,  53,  57, ...
  11,  15,  24,  28,  37,  41,  50,  54,  63,  67, ...
  14,  20,  31,  37,  48,  54,  65,  71,  82,  88, ...
  16,  22,  35,  41,  54,  60,  73,  79,  92,  98, ...
  19,  27,  42,  50,  65,  73,  88,  96, 111, 119, ...
  21,  29,  46,  54,  71,  79,  96, 104, 121, 129, ...
  24,  34,  53,  63,  82,  92, 111, 121, 140, 150, ...
  26,  36,  57,  67,  88,  98, 119, 129, 150, 160, ...
  ...
Note that, for example, the third row (or column) contains numbers that differ by 2 from multiples of 11 = 6*2 - 1, and the eighth row contains numbers that differ by 4 from multiples of 25 = 6*4 + 1.

The first and second rows are A047209 and A047336.

The diagonal is A062717, the numbers x for which 6*x + 1 is a perfect square.

a(m,n) = 6*floor((m+1)/2)*floor((n+1)/2) + ((-1)^n)*floor((m+1)/2) + ((-1)^m)*floor((n+1)/2);
matrix(7, 7, n, k, a(n, k)) \\ Michel Marcus, Jan 25 2019

a(m,n) = 6*floor((m+1)/2)*floor((n+1)/2) + ((-1)^n)*floor((m+1)/2) + ((-1)^m)*floor((n+1)/2), m,n >= 1.

cp's OEIS Frontend

A193053 a(n) = (14*n*(n+3) + (2*n-5)*(-1)^n + 21)/16.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A198442 Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (1,1,0) or (1,0,0).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Perl

Sage

Formula

Extensions

A175886 Numbers that are congruent to {1, 12} mod 13.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Magma

Mathematica

PARI

Formula

A175887 Numbers that are congruent to {1, 14} mod 15.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Magma

Mathematica

PARI

Formula

A195041 Concentric heptagonal numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Formula

A045472 Primes congruent to {1, 6} mod 7.

Views

Author

Keywords

Comments

Links

Crossrefs

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.