A062786 -id:A062786 - cp's OEIS frontend

A133292 Period 9: repeat [1, 1, 2, 4, 7, 2, 7, 4, 2].

1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2, 7, 4, 2, 1, 1, 2, 4, 7, 2

Offset: 0

Paul Curtz, Oct 17 2007

For n>0, digital roots of centered 10-gonal numbers (A062786). - Colin Barker, Jan 30 2015

For n>0, also the digital roots of central polygonal numbers (the Lazy Caterer's sequence) A000124. - Peter M. Chema, Sep 17 2016

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

Cf. A000124, A062786.

PadRight[{},111,{1,1,2,4,7,2,7,4,2}] (* Harvey P. Dale, Apr 29 2012 *)

a(n)=[1,1,2,4,7,2,7,4,2][n%9+1] \\ Charles R Greathouse IV, Jun 02 2011

Vec(-(2*x^8+4*x^7+7*x^6+2*x^5+7*x^4+4*x^3+2*x^2+x+1)/((x-1)*(x^2+x+1)*(x^6+x^3+1)) + O(x^100)) \\ Colin Barker, Apr 04 2015

G.f.: -(2*x^8+4*x^7+7*x^6+2*x^5+7*x^4+4*x^3+2*x^2+x+1) / ((x-1)*(x^2+x+1)*(x^6+x^3+1)). - Colin Barker, Apr 04 2015

a(n) = a(n-9). - Wesley Ivan Hurt, May 09 2022

A127561 Array T(n,k) = n^2+5nk+5*k^2 read downwards antidiagonals, n,k>=0.

Original entry on oeis.org

0, 5, 1, 20, 11, 4, 45, 31, 19, 9, 80, 61, 44, 29, 16, 125, 101, 79, 59, 41, 25, 180, 151, 124, 99, 76, 55, 36, 245, 211, 179, 149, 121, 95, 71, 49, 320, 281, 244, 209, 176, 145, 116, 89, 64, 405, 361, 319, 279, 241, 205, 171, 139, 109, 81, 500, 451, 404, 359, 316, 275, 236, 199, 164, 131

Offset: 0

Kenneth J Ramsey, Jan 18 2007, Feb 05 2007, Feb 06 2007

Lattice table of Fibonacci characteristic values from Wechsler's J determinant sequence A022344 uniquely position such that the row and column determine starting a,b values of a Fibonacci sequence having the same characteristic value.
A vector from (0,0) to any prime value P in the array does not pass through any other lattice point. If that vector is extended it passes through lattice points having successively the values 0, P*1^2, P*2^2, P*3^2, P*4^2 ... All primes ending in 1, 5 and 9 or the product thereof appear in the array, no prime ending in 3 or 7 appears in the array except in a square product which may be multiplied by a squarefree product of primes ending in 1, 5 or 9.
The table can be expanded by allowing negative arguments in the formula, but any positive value obtained can be expressed with nonnegative arguments.
The second row is the sequence A062786. The term in every succeeding row is 2* the term immediately above minus the next above term plus 2.
If the table is rearranged by shifting each column down by twice the column number, then the terms in second column would be equal to the row number squared plus the row number minus 1 and every succeeding term to the right would be equal to twice the left-hand term minus the next left-hand term minus 2.
It appears that any prime ending in 1,5, or 9 or any such prime times 5 appears only once in the table and that every power of such a prime or product thereof has one and only one nonnegative row and column position such that the row and column positions are coprime. A method for finding a coprime row and column position of the 2^n th power of any prime ending in 1,5,or 9, or of the product thereof, from the coprime row and column position of that prime or product is suggested by the discussion in the link titled "Wythoff Array, Pythagorean Triples, Primes".
It seems that if you stack the row and column positions of two numbers in the array that the determinant gives a column in which the product appears. Thus since the row and column position of 29 and 41 are 3,1 and 4,1 respectively then the product (41*29) appears in column 1*4 - 3*1 or column 1. The same value appears also in column -1 so 3*1-1*4 is a valid answer also. For our purposes however we choose the order that gives a positive value. Once the column number of the product is known it is easy to find the row number. There may be new determinant based math to find the row directly, but I don't know of any. It may happen that the row is negative, in which case the following transformation works a(r,c) = a(-r,c+r). Applied twice this transformation gives the original starting pair. I have yet to find any case in which one starts out with positive values for the row and column of each factor of a number appearing in the table and using the above determinant math cannot find positive values for the row and column of the product. I posted a few interesting results in the Cut-the-knot forum. Use the link given previously.

			T(0,1) = 5 because (0+2*1)^2 + 1*(0+2*1) - 1^2 = 5 and also because the Fibonacci sequence having the Horadam ID {a,b,1,1} with a = 0+2*1 and b = 1 has the characteristic value a^2 + b*a - b^2.
  0,  5, 20, 45, 80,125,180,245,320,405,500,... A033429
  1, 11, 31, 61,101,151,211,281,361,451,551,..  A062786
  4, 19, 44, 79,124,179,244,319,404,499,604,..  A134538
  9, 29, 59, 99,149,209,279,359,449,549,659,... A143198 (row 9)
 16, 41, 76,121,176,241,316,401,496,601,716,...
 25, 55, 95,145,205,275,355,445,545,655,775,..
 36, 71,116,171,236,311,396,491,596,711,836,...
 49, 89,139,199,269,349,439,539,649,769,899,...
 64,109,164,229,304,389,484,589,704,829,964,...
 81,131,191,261,341,431,531,641,761,891,1031,...
100,155,220,295,380,475,580,695,820,955,1100,...

K. J. Ramsey, Wythoff Array, Pythagorean triples, Primes

Cf. A035513, A022344.

T(a,b) = (a+2b)^2 + b(a+2b) - b^2.

A134462 Centered decagonal palindromic primes; or palindromic primes of the form 5k^2 + 5k + 1.

Original entry on oeis.org

11, 101, 151, 1598951, 1128512158211, 104216919612401, 107635959536701, 106906347292743609601, 165901968762984246868642489267869109561

Offset: 1

Alexander Adamchuk, Oct 26 2007

Sequence is the intersection of the palindromic primes = A002385 = {2, 3, 5, 7, 11, 101, 131, 151, ...} and the centered 10-gonal numbers = A062786 = {1, 11, 31, 61, 101, 151, ...}. Corresponding numbers k such that 5k^2 + 5k + 1 is a term of A134462 are listed in A134463 = {1, 4, 5, 565, 475081, ...}. Note that the first 4 terms of A134463 are palindromic as well.
a(9) > 10^25. - Donovan Johnson, Feb 13 2011
a(10) > 10^39. - Patrick De Geest, May 29 2021

Patrick De Geest, Palindromic Centered Polygonal Numbers
Eric Weisstein's World of Mathematics, Palindromic Prime
Wikipedia, Centered decagonal number.

Cf. A002385 = Palindromic primes.
Cf. A062786 = Centered 10-gonal numbers.
Cf. A090562 = Primes of the form 5k^2 + 5k + 1.
Cf. A090563 = Values of k such that 5k^2 + 5k + 1 is a prime.
Cf. A134463 = Values of k such that 5k^2 + 5k + 1 is a palindromic prime.

Do[ f=5k^2+5k+1; If[ PrimeQ[f] && FromDigits[ Reverse[ IntegerDigits[ f ] ] ] == f, Print[ f ] ], {k, 1, 500000} ]

More terms from Tomas J. Bulka (tbulka(AT)rodincoil.com), Aug 30 2009
a(8) from Donovan Johnson, Feb 13 2011
a(9) from Patrick De Geest, May 29 2021

A134463 Values of k such that 5k^2 + 5k + 1 is a palindromic prime.

Original entry on oeis.org

1, 4, 5, 565, 475081, 4565455, 4639740, 4623988479, 5760242508141202328

Offset: 1

Alexander Adamchuk, Oct 26 2007

Corresponding centered decagonal palindromic primes are 5k^2 + 5k + 1 = A134462 = {11, 101, 151, 1598951, 1128512158211, ...}. Note that the first 4 terms of A134463 are palindromic as well.
a(9) > 1414213562372. - Donovan Johnson, Feb 13 2011
a(10) > 14142135623730950488. - Patrick De Geest, May 29 2021

Patrick De Geest, Palindromic Centered Polygonal Numbers
Eric Weisstein's World of Mathematics, Palindromic Prime
Wikipedia, Centered decagonal number

Cf. A134462 = Centered decagonal palindromic primes; or palindromic primes of the form 5k^2 + 5k + 1.
Cf. A002385 = Palindromic primes.
Cf. A062786 = Centered 10-gonal numbers.
Cf. A090562 = Primes of the form 5k^2 + 5k + 1.
Cf. A090563 = Values of k such that 5k^2 + 5k + 1 is a prime.

Do[ f=5k^2+5k+1; If[ PrimeQ[f] && FromDigits[ Reverse[ IntegerDigits[ f ] ] ] == f, Print[ k ] ], {k, 1, 500000} ]

a(6), a(7) from D. S. McNeil, Mar 02 2009
a(8) from Donovan Johnson, Feb 13 2011
a(9) from Patrick De Geest, May 29 2021

A193248 Truncated dodecahedron, and truncated icosahedron with faces of centered polygons.

Original entry on oeis.org

1, 93, 455, 1267, 2709, 4961, 8203, 12615, 18377, 25669, 34671, 45563, 58525, 73737, 91379, 111631, 134673, 160685, 189847, 222339, 258341, 298033, 341595, 389207, 441049, 497301, 558143, 623755, 694317, 770009, 851011, 937503, 1029665, 1127677, 1231719

Offset: 1

Craig Ferguson, Jul 19 2011

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a truncated dodecahedron or truncated icosahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon of each face. [centered triangles (A005448) and centered decagons (A062786)] & [centered hexagons (A003215) and centered pentagons (A005891)] respectively.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Wikipedia, Tetrahedral number
Wikipedia, Triangular number
Wikipedia, Centered polygonal number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A003215, A005448, A005891, A062786.

[30*n^3-45*n^2+17*n-1: n in [1..50]]; // Vincenzo Librandi, Aug 30 2011

Table[30n^3-45n^2+17n-1,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,93,455,1267},40] (* Harvey P. Dale, Aug 28 2011 *)

vector(40, n, 30*n^3 - 45*n^2 + 17*n - 1) \\ G. C. Greubel, Nov 10 2018

a(n) = 30*n^3 - 45*n^2 + 17*n - 1.
G.f.: x*(1+x)*(x^2 + 88*x + 1) / (x-1)^4. - R. J. Mathar, Aug 26 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=1, a(2)=93, a(3)=455, a(4)=1267. - Harvey P. Dale, Aug 28 2011
E.g.f.: 1 - (1 - 2*x - 45*x^2 - 30*x^3)*exp(x). - G. C. Greubel, Nov 10 2018

A332243 Starhex honeycomb numbers: a(n) = 13 + 60n + 60n^2.

Original entry on oeis.org

13, 133, 373, 733, 1213, 1813, 2533, 3373, 4333, 5413, 6613, 7933, 9373, 10933, 12613, 14413, 16333, 18373, 20533, 22813, 25213, 27733, 30373, 33133, 36013, 39013, 42133, 45373, 48733, 52213, 55813, 59533, 63373, 67333, 71413, 75613, 79933, 84373

Offset: 0

John Elias, Feb 07 2020

			Example: a(2) = 13 + 60*2 + 60*2^2 = 373.
Illustration of initial terms:
.                               0
.                            0 0 0 0
.                             0 0 0
.                      0     0 0 0 0     0
.                   0 0 0 0 * * 0 * * 0 0 0 0
.                    0 0 0 * * * * * * 0 0 0
.                   0 0 0 0 * * 0 * * 0 0 0 0
.                      0 * * 0 0 0 0 * * 0
.                       * * * 0 0 0 * * *
.                      0 * * 0 0 0 0 * * 0
.                   0 0 0 0 * * 0 * * 0 0 0 0
.                    0 0 0 * * * * * * 0 0 0
.    0              0 0 0 0 * * 0 * * 0 0 0 0
. 0 * * 0              0     0 0 0 0     0
.  * 0 *                      0 0 0
. 0 * * 0                    0 0 0 0
.    0                          0
.
.    13                         133

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 20.

John Elias, Illustration of Initial Terms
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A003154, A003215, A062786.

Subsequence of A082369: cf. formula.

Array[12 (5 #^2 + 5 # + 1) + 1 &, 38, 0] (* Michael De Vlieger, Feb 07 2020 *)
LinearRecurrence[{3,-3,1},{13,133,373},40] (* Harvey P. Dale, Nov 18 2023 *)

A332243(n)=n*(n+1)*60+13 \\ M. F. Hasler, Jun 09 2023

a(n) = 12*(5*n*(n + 1) + 1) + 1.
From Stefano Spezia, Feb 07 2020: (Start)
O.g.f.: (13 + 94*x + 13*x^2)/(1 - x)^3.
E.g.f.: exp(x)*(13 + 120*x + 60*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-2) for n > 2. (End)
a(n) = A082369(A001844(n)). - M. F. Hasler, Jun 09 2023

A125239 Smallest prime divisor of 10T(n)+1 = 5n*(n+1)+1, where T(n) = 1 + 2 + ... + n.

Original entry on oeis.org

11, 31, 61, 101, 151, 211, 281, 19, 11, 19, 661, 11, 911, 1051, 1201, 1361, 1531, 29, 1901, 11, 2311, 2531, 11, 3001, 3251, 3511, 19, 31, 19, 4651, 11, 5281, 31, 11, 6301, 6661, 79, 7411, 29, 59, 79, 11, 9461, 9901, 11, 19, 29, 19, 12251, 41, 89, 13781, 11

Offset: 1

Nick Hobson, Nov 25 2006

All divisors of 10*T(n)+1 are congruent to 1 or -1 modulo 10; that is, they end in the decimal digit 1 or 9.

			10*T(9) + 1 = 5*9*10 + 1 = 451 = 11*41, so a(9) = 11.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Nick Hobson, Triangular Numbers.

Cf. A000217, A062786, A090562, A124989.

FactorInteger[#][[1,1]]&/@(10*Accumulate[Range[60]]+1) (* Harvey P. Dale, Dec 12 2011 *)

a(n) = if(n<1, 0, factor(5*n*(n+1)+1)[1,1])

A141534 Derived from the centered polygonal numbers: start with the first triangular number, then the sum of the first square number and the second triangular number, then the sum of first pentagonal number, the second square number and the third triangular number, and so on and so on...

Original entry on oeis.org

1, 4, 11, 26, 55, 105, 184, 301, 466, 690, 985, 1364, 1841, 2431, 3150, 4015, 5044, 6256, 7671, 9310, 11195, 13349, 15796, 18561, 21670, 25150, 29029, 33336, 38101, 43355, 49130, 55459, 62376, 69916, 78115, 87010, 96639, 107041, 118256, 130325

Offset: 1

Dan Graybill (clopen(AT)comcast.net), Aug 12 2008

Consider the array of triangular, square and centered polygonal numbers (irregular variant of A086272 and A086273):
 3   6  10  15  21  28  36  45  55  A000217
 4   9  16  25  36  49  64  81 100  A000290
 6  16  31  51  76 106 141 181 226  A005891
 7  19  37  61  91 127 169 217 271  A003215
 8  22  43  71 106 148 197 253 316  A069099
 9  25  49  81 121 169 225 289 361  A016754
10  28  55  91 136 190 253 325 406  A060544
11  31  61 101 151 211 281 361 451  A062786
12  34  67 111 166 232 309 397 496  A069125
13  37  73 121 181 253 337 433 541  A003154
14  40  79 131 196 274 365 469 586  A069126
15  43  85 141 211 295 393 505 631  A069127
etc. The sequence contains the antidiagonal sums of this array. - R. J. Mathar, Jun 05 2011
For comparison, the antidiagonal sums of A086270 are essentially A006522 starting at the 4th term. - R. J. Mathar, Sep 20 2008

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217.

a(n) = (n-1)*(n^3+11*n^2-38*n+120)/24, n>1. - R. J. Mathar, Sep 12 2008

G.f.: x*(1-x+x^2+x^3-x^5)/(1-x)^5. - Alexander R. Povolotsky, Jun 06 2011

A193253 Great rhombicosidodecahedron with faces of centered polygons.

Original entry on oeis.org

1, 183, 905, 2527, 5409, 9911, 16393, 25215, 36737, 51319, 69321, 91103, 117025, 147447, 182729, 223231, 269313, 321335, 379657, 444639, 516641, 596023, 683145, 778367, 882049, 994551, 1116233, 1247455, 1388577, 1539959, 1701961, 1874943, 2059265, 2255287

Offset: 1

Craig Ferguson, Jul 19 2011

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a great rhombicosidodecahedron. Each iteration requires the addition of n-2 edges and n-1 vertices to complete the centered polygon of each face.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000.
OEIS Wiki, (Centered_polygons) pyramidal numbers.
Eric W. Weisstein, MathWorld: Great Rhombicosidodecahedron.
Wikipedia, Tetrahedral number.
Wikipedia, Triangular number.
Wikipedia, Centered polygonal number.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001844 (centered squares), A062786 (centered decagons), and A003215 (centered hexagons).

=60*ROW()^3-90*ROW()^2+32*ROW()-1 fill down  to desired size.

[60*n^3-90*n^2+32*n-1: n in [1..40]] // Vincenzo Librandi, Feb 18 2012

LinearRecurrence[{4, -6, 4, -1}, {1, 183, 905, 2527}, 50] (* Vincenzo Librandi, Feb 18 2012 *)
a[n_]:=60*n^3 - 90*n^2 + 32*n - 1 ; Array[a, 50] (* or *)
CoefficientList[Series[(1 + x)*(1 + 178*x + x^2)/(1 - x)^4 , {x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)

a(n)=60*n^3-90*n^2+32*n-1 \\ Charles R Greathouse IV, Feb 12 2012

a(n) = 60*n^3 - 90*n^2 + 32*n - 1.

G.f.: x*(1 + 179*x + 179*x^2 + x^3)/(1-x)^4 = x*(1+x)*(1 + 178*x + x^2)/(1-x)^4. - Colin Barker, Feb 12 2012

A193448 a(n) = 4(5n^2 - 5*n + 1).

Original entry on oeis.org

4, 44, 124, 244, 404, 604, 844, 1124, 1444, 1804, 2204, 2644, 3124, 3644, 4204, 4804, 5444, 6124, 6844, 7604, 8404, 9244, 10124, 11044, 12004, 13004, 14044, 15124, 16244, 17404, 18604, 19844, 21124, 22444, 23804, 25204, 26644, 28124, 29644, 31204, 32804

Offset: 1

Giovanni Teofilatto, Jul 26 2011

The natural numbers of the form 5*n^2-1, with n odd. See also A158491 for the cases where n is even. - Giovanni Teofilatto, Oct 10 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A062786, A158491.

[4*(5*n^2-5*n+1): n in [1..50]]; // Vincenzo Librandi, Aug 30 2011

A193448:=n->4*(5*n^2 - 5*n + 1): seq(A193448(n), n=1..50); # Wesley Ivan Hurt, Nov 21 2015

Table[4*(5*n^2 - 5*n + 1), {n, 50}] (* Wesley Ivan Hurt, Nov 21 2015 *)

a(n) = 4*(5*n^2 - 5*n + 1) \\ Anders Hellström, Nov 21 2015

a(n) = 4*A062786(n).
G.f.: -4*x*(1+8*x+x^2) / (x-1)^3. - R. J. Mathar, Aug 26 2011
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>3. - Wesley Ivan Hurt, Nov 21 2015

cp's OEIS Frontend

A133292 Period 9: repeat [1, 1, 2, 4, 7, 2, 7, 4, 2].

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A127561 Array T(n,k) = n^2+5*n*k+5*k^2 read downwards antidiagonals, n,k>=0.

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A134462 Centered decagonal palindromic primes; or palindromic primes of the form 5k^2 + 5k + 1.

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A134463 Values of k such that 5k^2 + 5k + 1 is a palindromic prime.

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A193248 Truncated dodecahedron, and truncated icosahedron with faces of centered polygons.

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A332243 Starhex honeycomb numbers: a(n) = 13 + 60*n + 60*n^2.

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A125239 Smallest prime divisor of 10*T(n)+1 = 5*n*(n+1)+1, where T(n) = 1 + 2 + ... + n.

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A141534 Derived from the centered polygonal numbers: start with the first triangular number, then the sum of the first square number and the second triangular number, then the sum of first pentagonal number, the second square number and the third triangular number, and so on and so on...

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A127561 Array T(n,k) = n^2+5nk+5*k^2 read downwards antidiagonals, n,k>=0.

A332243 Starhex honeycomb numbers: a(n) = 13 + 60n + 60n^2.

A125239 Smallest prime divisor of 10T(n)+1 = 5n*(n+1)+1, where T(n) = 1 + 2 + ... + n.

A193448 a(n) = 4(5n^2 - 5*n + 1).