John Rafael M. Antalan - cp's OEIS frontend

User: John Rafael M. Antalan

John Rafael M. Antalan has authored 3 sequences.

A270889 Integers n such that the circular graph C_n has a square size deficiency.

3, 6, 27, 150, 867, 5046, 29403, 171366, 998787, 5821350, 33929307, 197754486, 1152597603, 6717831126, 39154389147, 228208503750, 1330096633347, 7752371296326, 45184131144603, 263352415571286, 1534930362283107, 8946229758127350, 52142448186480987, 303908459360758566, 1771308307978070403

Offset: 0

John Rafael M. Antalan, Mar 25 2016

Define the size deficiency of a graph G as the number of edges needed to complete G. If G is a cycle graph C_n, this sequence gives the values of n for which C_n has a size deficiency which is a perfect square.

Colin Barker, Table of n, a(n) for n = 0..1000
J. R. M. Antalan and I. F. Callano, On the Size Deficiency of Cycle Graphs and Some Integer Sequences, Asian Journal of Mathematics and Computer Research, 11 (3) (2016),192-200.
Mathematics Stack Exchange, Solving the nonlinear Diophantine equation x2-3x=2y^2
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

a[0] = 3; a[1] = 6; a[n_] := a[n] = 6 a[n - 1] - a[n - 2] - 6; Table[a@ n, {n, 0, 24}] (* Michael De Vlieger, Mar 25 2016 *)
LinearRecurrence[{7,-7,1},{3,6,27},30] (* Harvey P. Dale, Jan 23 2019 *)

is(n)=issquare(n*(n-3)/2) \\ Charles R Greathouse IV, Mar 25 2016

a(n)=([0,1,0;0,0,1;1,-7,7]^n*[3;6;27])[1,1] \\ Charles R Greathouse IV, Mar 25 2016

Vec(3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, Apr 03 2016

a(n+2) = 6*a(n+1) - a(n) - 6; a(0) = 3 , a(1) = 6.
G.f.: 3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)). - Joerg Arndt, Mar 25 2016
a(n) = 3 * A055997(n+1). - Joerg Arndt, Mar 25 2016
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2. - Colin Barker, Apr 03 2016
a(n) = 3*(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4. - Colin Barker, Apr 03 2016

A261654 Lead almost-Pythagorean triples generated by primitive Pythagorean triples of the form (2i-1, 2i^2-2i, 2i^2-2i+1), i >= 2.

Original entry on oeis.org

4, 7, 8, 6, 17, 18, 8, 31, 32, 10, 49, 50, 12, 71, 72, 14, 97, 98, 16, 127, 128, 18, 161, 162, 20, 199, 200, 22, 241, 242, 24, 287, 288, 26, 337, 338, 28, 391, 392, 30, 449, 450, 32, 511, 512, 34, 577, 578

Offset: 1

John Rafael M. Antalan, Aug 30 2015

A set of ordered triple (x,y,z) that satisfies the equation x^2 + y^2 = z^2 + 1 is called an almost-Pythagorean triple (APT).
The triples (x,y,z)=[(2i-1)k+1,(2i^2-2i)k+(2i-1),(2i^2-2i+1)k+(2i-1)] and (x',y',z')=[(2i-1)k+(2i-2),(2i^2-2i)k+(2i^2-4i+1),(2i^2-2i+1)k+(2i^2-4i+2)] are APTs for all integers k and i >= 2.
Note that in terms of components, (x,y,z) < (x',y',z').
Setting k=1 in the first expression gives the terms of this sequence.

			When k=1 and i=2 the formula for (x,y,z) gives the Lead APT (4,7,8).
First rows are:
   4,  7,  8;
   6, 17, 18;
   8, 31, 32;
  10, 49, 50;
  12, 71, 72;
  14, 97, 98;
  ...

John Rafael M. Antalan, Mark D. Tomenes, A Note on Generating Almost Pythagorean Triples, arXiv:1508.07562 [math.NT], 2015.
O. Frink, Almost Pythagorean Triples, Mathematics Magazine, Vol.60, No.4, (1987), pp.234-236.

For the 3 columns, cf. A005843, A056220, A001105.

xyz[i_] := {2i, 2i^2-1, 2i^2};
Array[xyz, 16, 2] // Flatten (* Jean-François Alcover, Feb 02 2019 *)

tabf(nn) = for (i=2, nn, print(2*i, ", ", 2*i^2-1, ", ", 2*i^2)); \\ Michel Marcus, Aug 31 2015

(x,y,z) = [(2i-1)k+1,(2i^2-2i)k+(2i-1),(2i^2-2i+1)k+(2i-1)], with i>=2 and k=1.

A256704 Palindromes of the form 4n + 1 that are divisible by 5.

Original entry on oeis.org

5, 505, 525, 545, 565, 585, 5005, 5225, 5445, 5665, 5885, 50005, 50105, 50205, 50305, 50405, 50505, 50605, 50705, 50805, 50905, 52025, 52125, 52225, 52325, 52425, 52525, 52625, 52725, 52825, 52925, 54045, 54145, 54245, 54345, 54445, 54545, 54645, 54745, 54845

Offset: 1

John Rafael M. Antalan, Apr 08 2015

The sequence is infinite.
Creation of the Sequence. The sequence was generated based on the following known results for primitive Pythagorean triples/triangles (PPTs).
Let c represent the hypotenuse of a PPT and a, b be its sides. Without loss of generality, let a be the odd side and b be the even side.
(a) c takes the form 4n+1.
(b) One of a,b,c is divisible by 5.
Let c be divisible by 5. Thus, this sequence represents the palindromic hypotenuse of a PPT that is divisible by 5.
With the exception of 5, the last two digits of each term of this sequence are 05, 25, 45, 65 or 85. Thus, since the terms are palindromes, the first two digits of each term of the sequence are 50, 52, 54, 56 or 58.
Proof. If k is an element of the sequence, then its form must be 4n+1 for some integer n. This means that k-1=4n that is, k-1 is divisible by 4. Since an integer is divisible by 4 if and only if the number composed of its last two digits is divisible by 4, it must be that the number composed of the last two digits of k-1 is divisible by 4. With this we get the desired result.
Except for the terms having one or three digits, the terms of the sequence take the following forms:
(a) 50p05
(b) 52p25
(c) 54p45
(d) 56p65
(e) 58p85
where p is a palindromic number with leading zero digit(s) allowed.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
John Rafael M. Antalan and Richard P. Tagle, Numeric Palindromes in Primitive and Non-Primitive Pythagorean Triples, arXiv:1502.06760 [math.NT], 2015.

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Select[Select[4 Range[100000] + 1, palQ], Mod[#, 5] == 0 &] (* Michael De Vlieger, Apr 08 2015 *)
Select[5*Range[12000],PalindromeQ[#]&&Mod[#,4]==1&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 09 2018 *)

ispal(n) = (d = digits(n)) && (Vecrev(d) == d);
isok(n) = !(n % 5) && !((n-1) % 4) && ispal(n); \\ Michel Marcus, Apr 08 2015

More terms from Michel Marcus, Apr 08 2015

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User: John Rafael M. Antalan

A270889 Integers n such that the circular graph C_n has a square size deficiency.

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A261654 Lead almost-Pythagorean triples generated by primitive Pythagorean triples of the form (2i-1, 2i^2-2i, 2i^2-2i+1), i >= 2.

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A256704 Palindromes of the form 4n + 1 that are divisible by 5.

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