A175885 -id:A175885 - cp's OEIS frontend

A219257 Numbers k such that 11*k+1 is a square.

0, 9, 13, 40, 48, 93, 105, 168, 184, 265, 285, 384, 408, 525, 553, 688, 720, 873, 909, 1080, 1120, 1309, 1353, 1560, 1608, 1833, 1885, 2128, 2184, 2445, 2505, 2784, 2848, 3145, 3213, 3528, 3600, 3933, 4009, 4360, 4440, 4809, 4893, 5280, 5368, 5773, 5865

Offset: 1

Bruno Berselli, Nov 16 2012

Equivalently, numbers of the form m*(11*m+2), where m = 0,-1,1,-2,2,-3,3,...

Also, integer values of h*(h+2)/11.

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

Cf. numbers k such that h*k+1 is a square: A005563 (h=1), A046092 (h=2), A001082 (h=3), A002378 (h=4), A036666 (h=5), A062717 (h=6), A132354 (h=7), A000217 (h=8), A132355 (h=9), A132356 (h=10), A152749 (h=12), A219389 (h=13), A219390 (h=14), A204221 (h=15), A074378 (h=16), A219394 (h=17), A219395 (h=18), A219396 (h=19), A219190 (h=20), A219391 (h=21), A219392 (h=22), A219393 (h=23), A001318 (h=24), A219259 (h=25), A217441 (h=26), A219258 (h=27), A219191 (h=28).

Cf. A175885 (square roots of 11*a(n)+1).

[n: n in [0..7000] | IsSquare(11*n+1)];

I:=[0,9,13,40,48]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

Select[Range[0, 7000], IntegerQ[Sqrt[11 # + 1]] &]
CoefficientList[Series[x (9 + 4 x + 9 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

G.f.: x^2*(9+4*x+9*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (22*n*(n-1)+7*(-1)^n*(2*n-1)-1)/8 + 1 = (1/176)*(22*n+7*(-1)^n-15)*(22*n+7*(-1)^n-7).
Sum_{n>=2} 1/a(n) = 11/4 - cot(2*Pi/11)*Pi/2. - Amiram Eldar, Mar 15 2022

A091998 Numbers that are congruent to {1, 11} mod 12.

Original entry on oeis.org

1, 11, 13, 23, 25, 35, 37, 47, 49, 59, 61, 71, 73, 83, 85, 95, 97, 107, 109, 119, 121, 131, 133, 143, 145, 155, 157, 167, 169, 179, 181, 191, 193, 203, 205, 215, 217, 227, 229, 239, 241, 251, 253, 263, 265, 275, 277, 287, 289, 299, 301, 311, 313, 323, 325, 335

Offset: 1

Ray Chandler, Feb 21 2004

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 and n in A000027), then ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 12). Also a(n)^2 - 1 == 0 (mod 24).

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

First row of A092260.
Cf. A175885 (n == 1 or 10 (mod 11)), A175886 (n == 1 or 12 (mod 13)).
Cf. A097933 (primes), A195143 (partial sums).
Cf. A000217, A005408, A047209, A007310, A047336, A047522, A056020, A090771, A113801, A175887, A188887.

a091998 n = a091998_list !! (n-1)
a091998_list = 1 : 11 : map (+ 12) a091998_list
-- Reinhard Zumkeller, Jan 07 2012

[ n: n in [1..350] | n mod 12 eq 1 or n mod 12 eq 11 ];

LinearRecurrence[{1,1,-1},{1,11,13},100] (* Harvey P. Dale, Jul 26 2017 *)

is(n)=n=n%12;n==11 || n==1 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 12*n - a(n-1) - 12 (with a(1)=1). - Vincenzo Librandi, Nov 16 2010
a(n) = 6*n + 2*(-1)^n - 3.
G.f.: x*(1+10*x+x^2)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 12 for n > 2.
a(n) = 12*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2 + sqrt(3))*Pi/12. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + (6*x - 3)*exp(x) + 2*exp(-x). - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2 + sqrt(3)) = 2*cos(Pi/12) (A188887).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/3)*cos(Pi/12). (End)

Formulae and comment added by Bruno Berselli, Nov 17 2010 - Nov 18 2010

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)

A195043 Concentric 11-gonal numbers.

Original entry on oeis.org

0, 1, 11, 23, 44, 67, 99, 133, 176, 221, 275, 331, 396, 463, 539, 617, 704, 793, 891, 991, 1100, 1211, 1331, 1453, 1584, 1717, 1859, 2003, 2156, 2311, 2475, 2641, 2816, 2993, 3179, 3367, 3564, 3763, 3971, 4181, 4400, 4621, 4851, 5083, 5324, 5567

Offset: 0

Omar E. Pol, Sep 27 2011

Also concentric hendecagonal numbers. A033584 and A069173 interleaved.

Partial sums of A175885. - 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 11 of A195040.

Cf. A032527, A032528, A033584, A069173, A175885, A195042, A195142, A195143, A195045.

a195043 n = a195043_list !! n
a195043_list = scanl (+) 0 a175885_list
-- Reinhard Zumkeller, Jan 07 2012

[11*n^2/4+7*((-1)^n-1)/8: n in [0..50]]; // Vincenzo Librandi, Sep 30 2011

LinearRecurrence[{2,0,-2,1},{0,1,11,23},50] (* Harvey P. Dale, May 20 2019 *)

Vec(-x*(x^2+9*x+1)/((x-1)^3*(x+1)) + O(x^100)) \\ Colin Barker, Sep 15 2013

a(n) = 11*n^2/4 + 7*((-1)^n - 1)/8.
a(n) = -a(n-1) + A069125(n). - Vincenzo Librandi, Sep 30 2011
From Colin Barker, Sep 15 2013: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: -x*(x^2+9*x+1) / ((x-1)^3*(x+1)). (End)
Sum_{n>=1} 1/a(n) = Pi^2/66 + tan(sqrt(7/11)*Pi/2)*Pi/sqrt(77). - Amiram Eldar, Jan 16 2023

A267755 Expansion of (1 + 2x + x^2 + x^3 + 4x^4 + 2*x^5)/(1 - x - x^5 + x^6).

Original entry on oeis.org

1, 3, 4, 5, 9, 12, 14, 15, 16, 20, 23, 25, 26, 27, 31, 34, 36, 37, 38, 42, 45, 47, 48, 49, 53, 56, 58, 59, 60, 64, 67, 69, 70, 71, 75, 78, 80, 81, 82, 86, 89, 91, 92, 93, 97, 100, 102, 103, 104, 108, 111, 113, 114, 115, 119, 122, 124, 125, 126, 130, 133, 135, 136, 137

Offset: 0

Bruno Berselli, Jan 20 2016

(m^k-1)/11 is a nonnegative integer when
. m is a member of this sequence and k is an odd multiple of 5 (A017329),
. m is a member of A017401 and k is odd but not multiple of 5 (A045572),
. m is a member of A175885 and k is even but not multiple of 5 (A217562),
. m is a member of A160542 and k is a positive multiple of 10 (A008592),
apart from the trivial case in which k=0.
Also, numbers that are congruent to {1, 3, 4, 5, 9} mod 11. Therefore, the product of two terms belongs to the sequence.
Union of this sequence and A267541 is A160542.
a(n) is prime for n = 1, 3, 10, 14, 17, 21, 24, 27, 30, 33, 40, 44, 47, ...

			From the linear recurrence:
(-A267541) ..., -13, -10, -8, -7, -6, -2, 1, 3, 4, 5, 9, 12, ... (A267755)

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

Cf. A160542, A017329, A267541.

Related sequences (see the first comment): A017401, A160542, A175885.

m:=70; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+x^2+x^3+4*x^4+2*x^5)/(1-x-x^5+x^6)));

I:=[1,3,4,5,9,12]; [n le 6 select I[n] else Self(n-1)+Self(n-5)-Self(n-6): n in [1..70]]; // Vincenzo Librandi, Jan 21 2016

gf := (1 + 2*x + x^2 + x^3 + 4*x^4 + 2*x^5)/(1 - x - x^5 + x^6): deg := 64: series(gf, x, deg): seq(coeff(%, x, n), n=0..deg-1); # Peter Luschny, Jan 21 2016

CoefficientList[Series[(1 + 2 x + x^2 + x^3 + 4 x^4 + 2 x^5)/(1 - x - x^5 + x^6), {x, 0, 70}], x]
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {1, 3, 4, 5, 9, 12}, 70]
Select[Range[140], MemberQ[{1, 3, 4, 5, 9}, Mod[#, 11]]&]

Vec((1+2*x+x^2+x^3+4*x^4+2*x^5)/(1-x-x^5+x^6)+O(x^70))

gf = (1 + 2*x + x^2 + x^3 + 4*x^4 + 2*x^5)/(1 - x - x^5 + x^6)
print(taylor(gf, x, 0, 63).list()) # Peter Luschny, Jan 21 2016

G.f.: (1 + 2*x + x^2 + x^3 + 4*x^4 + 2*x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6).
a(-n) = -A267541(n-1).
a(n) = n + 1 + 2*floor(n/5) + 3*floor((n+1)/5) + floor((n+4)/5). - Ridouane Oudra, Sep 06 2023

A320170 Every pair of consecutive numbers sums to a palindrome, starting with a(0)=1; a(1)=10, and taking a(n) = smallest number > a(n-2).

Original entry on oeis.org

1, 10, 12, 21, 23, 32, 34, 43, 45, 54, 47, 64, 57, 74, 67, 84, 77, 94, 87, 104, 98, 114, 108, 124, 118, 134, 128, 144, 138, 154, 149, 164, 159, 174, 169, 184, 179, 194, 189, 204, 200, 214, 210, 224, 220, 234, 230, 244, 240, 254, 251, 264, 261, 274, 271

Offset: 0

Jim Singh, Oct 07 2018

Cf. A062932.

For n < 10, equals A061870 and A175885.

Nest[Append[#, Block[{k = #[[-2]] + 1}, While[! PalindromeQ[#[[-1]] + k], k++]; k]] &, {1, 10}, 53] (* Michael De Vlieger, Oct 10 2018 *)

CurrentNum=10
PreviousNum=1
NewNum=0
def NextPalindrome(StartNum):
    FoundNext=0
    t=0
    n=0
    Number=0
    Lastdigit=0
    while FoundNext<=PreviousNum:
        n=n+1
        t=StartNum+n
        Number=t
        Reverse=0
        while Number>0:
            Lastdigit=Number%10
            Reverse=(Reverse*10)+Lastdigit
            Number=Number//10
        if t==Reverse:
            FoundNext=n
    return n
print(1)
print(10)
for x in range(50):
    NewNum=NextPalindrome(CurrentNum)
    print(NewNum)
    PreviousNum=CurrentNum
    CurrentNum=NewNum

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A219257 Numbers k such that 11*k+1 is a square.

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A091998 Numbers that are congruent to {1, 11} mod 12.

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

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

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A195043 Concentric 11-gonal numbers.

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A267755 Expansion of (1 + 2*x + x^2 + x^3 + 4*x^4 + 2*x^5)/(1 - x - x^5 + x^6).

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A267755 Expansion of (1 + 2x + x^2 + x^3 + 4x^4 + 2*x^5)/(1 - x - x^5 + x^6).