A249079 a(n) = 29*n + floor( n/29 ) + 0^( 1-floor( (14+(n mod 29))/29 ) ).

0, 29, 58, 87, 116, 145, 174, 203, 232, 261, 290, 319, 348, 377, 406, 436, 465, 494, 523, 552, 581, 610, 639, 668, 697, 726, 755, 784, 813, 842, 871, 900, 929, 958, 987, 1016, 1045, 1074, 1103, 1132, 1161, 1190, 1219, 1248, 1278, 1307, 1336

Offset: 0

Author

Karl V. Keller, Jr., Oct 20 2014

Keywords

• nonn

Comments

This is an approximation to A004942 (Nearest integer to n*phi^7, where phi is the golden ratio, A001622).

Examples

			n= 0, 29*n+floor(0.0) +0^(1-floor(0.48))=    0 +0 +0 =    0 (n/29=0,0^1=0).
n=14, 29*n+floor(0.48)+0^(1-floor(0.97))=  406 +0 +0 =  406 (0^1=0).
n=15, 29*n+floor(0.52)+0^(1-floor(1.0)) =  435 +0 +1 =  436 (0^0=1).
n=28, 29*n+floor(0.97)+0^(1-floor(1.45))=  812 +0 +1 =  813 (0^0=1).
n=29, 29*n+floor(1.0) +0^(1-floor(0.48))=  841 +1 +0 =  842 (n/29*1,0^1=0).
n=43, 29*n+floor(1.48)+0^(1-floor(0.97))= 1247 +1 +0 = 1248 (0^1=0).
n=44, 29*n+floor(1.52)+0^(1-floor(1.0)) = 1276 +1 +1 = 1278 (0^0=1).
n=58, 29*n+floor(2.0) +0^(1-floor(0.48))= 1682 +2 +0 = 1684 (n/29*2,0^1=0).
n=85, 29*n+floor(2.93)+0^(1-floor(1.41))= 2465 +2 +1 = 2468 (0^0=1).
n=86, 29*n+floor(2.97)+0^(1-floor(1.45))= 2494 +2 +1 = 2497 (0^0=1).
n=87, 29*n+floor(3.0) +0^(1-floor(0.48))= 2523 +3 +0 = 2526 (n/29*3,0^0=0).
		

Links

• Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
• Ron Knott, Fibonacci numbers
• Eric Weisstein's World of Mathematics, Golden Ratio
• Wikipedia, Golden ratio
• Index entries for linear recurrences with constant coefficients, signature (1, 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, 1, -1).

Crossrefs

Cf. A001622 (phi), A195819 (29*n).

Cf. A004942 (round(n*phi^7)), A004922 (floor(n*phi^7)), A004962 (ceiling(n*phi^7)).

Programs

Magma

[29*n + Floor(n/29) + 0^(1-Floor((14+(n mod 29))/29)) : n in [0..50]]; // Vincenzo Librandi, Nov 05 2014

PARI

a(n) = 29*n + n\29 + 0^(1 - (14+(n % 29))\29); \\ Michel Marcus, Oct 25 2014

Python

for n in range(101):
    print(29*n+n//29+0**(1-(14+n%29)//29), end=', ')

Python

def A249079(n):
    a, b = divmod(n,29)
    return 29*n+a+int(b>=15) # Chai Wah Wu, Jul 27 2022